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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((8x-x^(2))))` |
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Answer» Correct Answer - `sin^(-1)((x-4)/(4))` `8x-x^(2)=-(x^(2)-8x+16-16)=-(x^(2)-8x+16)+16` `=4^(2)-(x-4)^(2)` |
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| 102. |
फलनों का x के सापेक्ष समाकलन कीजिए - `sqrt((5-4x-x^(2)))` |
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Answer» Correct Answer - `(1)/(2)[(x+2)sqrt((5-4x-x^(2)))+9sin^(-1)((x+2)/(3))]` `int(2x+1)sqrt(x^(2)+2x+5)dx` `=int[(2x+2)-1]sqrt(x^(2)+2x+5)dx` `=int(2x+2)sqrt(x^(2)+2x+5)dx-intsqrt(x^(2)+2x+5)dx` अब माना `x^(2)+2+5 = t rArr (2x+2)dx=dt` |
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| 103. |
फलनों का x के सापेक्ष समाकलन कीजिए - `sqrt((2x^(2)+3x+4))` |
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Answer» Correct Answer - `(4x+3)/(8)sqrt((2x^(2)+3x+4))+(23sqrt2)/(32)log[(x+(3)/(4))+sqrt(x^(2)+(3)/(2)x+2)]` `2x^(2)+3x+4=2(x^(2)+(3)/(2)x+2)` `=2(x^(2)+(3)/(2)x+(9)/(16)-(9)/(16)+2)` `=2(x^(2)+(3)/(2)x+(9)/(16))+2xx(23)/(16)` `=2[(x+(3)/(4))^(2)+(23)/(16)]` |
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| 104. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(3x-2)sqrt(x^(2)+x+1)` |
| Answer» `(x^(2)+x+1)^(3//2)-(7(2x+1))/(8)-(21)/(16)log[(x+(1)/(2))+sqrt((x^(2)+x+1))]` | |
| 105. |
फलनों का x के सापेक्ष समाकलन कीजिए - `sqrt(15-2x-x^(2))` |
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Answer» Correct Answer - `(1)/(2)[(x+1)sqrt((15-2x-x^(2)))+16 sin^(-1)((x+1)/(4))]` `15-2x-x^(2)=(4)^(2)-(x+1)^(2)` |
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| 106. |
फलनों का समाकलन कीजिए - `(1)/(a^(2)x^(2)-b^(2))` |
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Answer» Correct Answer - `(1)/(2ab)log((ax-b)/(ax+b))` `int(1)/(a^(2)x^(2)-b^(2))dx=int(1)/((ax)^(2)-b^(2))dx` अब माना `ax=t rArr adx= dt` |
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| 107. |
फलनों का समाकलन कीजिए - `(1)/(2x^(2)+5x+3)` |
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Answer» Correct Answer - `log((x+1)/(2x+3))` `2x^(2)+5x+3=(x+1)(2x+3)` `therefore int(dx)/(2x^(2)+5x+)=int(dx)/((x+1)(2x+3))=int[(1)/(x+1)-(2)/(2x+3)]dx` `=log(x+1)-(2)/(2)log(2x+3)+c` `=log((x+1)/(2x+3))+c` |
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| 108. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x+1)/(sqrt((x^(2)+4x+1)))` |
| Answer» `-(1)/(5)sqrt((4+8x-5x^(2)))+(9)/(5sqrt5)sin^(-1)((5x-4)/(6))` | |
| 109. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x+2)/(sqrt((4x-x^(2))))` |
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Answer» Correct Answer - `-sqrt(4x-x^(2))+4sin^(-1)((x-2)/(2))` `(x+2)/(sqrt(4x-x^(2)))=-(1)/(2).(-2x-4)/(sqrt(4x-x^(2)))` `=-(1)/(2).((-2x-4)+4-4)/(sqrt(4x-x^(2)))=(1)/(2).((-2x-4)-8)/(sqrt(4x-x^(2)))` `=-(1)/(2).((-2x-4))/(sqrt(4x-x^(2)))+(4)/(sqrt(4x-x^(2)))` `4x-x^(2)=2^(2)-(x-2)^(2)` |
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| 110. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x+8)/(sqrt(x^(2)+2x+5))` |
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Answer» Correct Answer - `sqrt((x^(2)+2x+5))+7log[(x+1)+sqrt((x^(2)+2x+5))]` `(x+8)/(sqrt(x^(2)+2x+5))=(1)/(2).((2x+2))/(sqrt(x^(2)+2x+5))+7 xx(1)/(sqrt(x^(2)+2x+5))` |
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| 111. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((12-x^(2)-4x)))` |
| Answer» `sin^(-1)((x+2)/(4))` | |
| 112. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x-1)/(sqrt((x^(2)-x+1)))` |
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Answer» Correct Answer - `sqrt((x^(2)-x+1))-(1)/(2)log[(x-(1)/(2))+sqrt((x^(2)-x+1))]` `(x-1)/(sqrt(x^(2)-x+1))=(1)/(2)int((2x-1)-1)/(sqrt(x^(2)-x+1))=(1)/(2)((2x-1)/(sqrt(x^(2)-x+1)))-(1)/(2)(1)/(sqrt(x^(2)-x+1))` |
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| 113. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((2+x-3x^(2))))` |
| Answer» `(1)/(sqrt3)sin^(-1)((6x-1)/(5))` | |
| 114. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(2x+1)sqrt(x^(2)+2x+5)` |
| Answer» `(2)/(3)(x^(2)+2x+5)^(3//2)-(1)/(2)(x+1)sqrt((x^(2)+2x+5))+4 log {(x+1)+sqrt((x^(2)+2x+5))+4log (x+1)+sqrt((x^(2)+2x+5))}` | |
| 115. |
फलनों का समाकलन कीजिए - `(3x+1)/(2x^(2)-x+3)` |
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Answer» Correct Answer - `(3)/(4)log(2x^(2)+x+1)+(1)/(2sqrt7)tan^(-1).(4x+1)/(sqrt7)` `(x+1)/(x^(2)+4x+5)=(2x+4)/(2(x^(2)+4x+5))-(1)/((x^(2)+4x+5))` `=(1)/(2).(2x+4)/((x^(2)+4x+5))-(1)/((x+2)^(2)+1)` |
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| 116. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt(1-4x-2x^(2)))` |
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Answer» Correct Answer - `(1)/(sqrt2)sin^(-1)[((x+1)sqrt2)/(sqrt3)]` `1-4x-2x^(2)=-(2x^(2)+4x-1)=-(2x^(2)+4x-3+2)` `=-2(x^(2)+2x+1)+3=2[(sqrt((3)/(2)))^(2)-(x+1)^(2)]` |
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| 117. |
फलनों का समाकलन कीजिए - `(x)/((x-1)^(2)(x-2))` |
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Answer» Correct Answer - `2log((x-2)/(x-1))+(1)/(x-1)` `(x)/((x-1)^(2)(x-2))=(2)/((x-2))-(2)/((x-1))-(1)/((x-1)^(2))` |
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| 118. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x+1)/(sqrt(4-5x-x^(2)))` |
| Answer» `-sqrt((4+5x-x^(2)))+(7)/(2)sin^(-1)((2x-5)/(sqrt(41)))` | |
| 119. |
फलनों का समाकलन कीजिए - `(1)/(1-x-2x^(2))` |
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Answer» Correct Answer - `(1)/(3)log((x+1)/(1-2x))` `int(1)/(1-x-2x^(2))=int[(1)/(3(x+1))+(2)/(3(1-2x))]dx` |
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| 120. |
फलनों का समाकलन कीजिए - `(1)/(x(x^(n)+1))` |
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Answer» Correct Answer - `(1)/(n)log((x^(n))/(1+x(n)))` `int(1)/(x(x^(n)+1))dx=int(x^(n-1))/(x.x^(n-1)(x^(n)+1))dx=int(x^(n-1))/(x^(n)(x^(n)+1))dx` अब माना `x^(n)=t` |
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| 121. |
`int(cosx)/(3-4 sin^(2)x)dx` का मान ज्ञात कीजिए । |
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Answer» माना `" "2sin x= t " "rArr" "2 cos x dx =dt` `therefore int(cosx)/(3-4 sin^(2)x)dx=(1)/(2)int(dt)/((sqrt3)^(2)-t^(2))` `=(1)/(2)xx(1)/(2sqrt3)log((sqrt3+t)/(sqrt3-t))` `=(1)/(4sqrt3)log((sqrt3 +2sinx)/(sqrt3-2 sinx))` |
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| 122. |
`int(3)/((x-2)(x+1))dx` का मान ज्ञात कीजिए । |
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Answer» माना `(3x)/((x-2)(x+1))=(A)/((x-2))+(B)/((x+1))` `=(A(x+1)+B(x-1))/((x-2)(x+1))` `rArr" "3x=x(A+B)+(A-2B)` दोनों और x के समान पदों के गुणकों की तुलना करने पर `A+B=3` व `" "A-2B=0` उपरोक्त समीकरणों को हल करने पर,, `A=2, B=1` इसलिए `int(3x)/((x-2)(x+1))dx=int(2)/((x-2))dx+int(1)/((x+1)dx` `=2log (x-2)+log(x+1)+c` `=log(x-2)^(2).(x+1)+c` |
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| 123. |
`int(dx)/(sqrt(7-6x-x^(2)))` |
| Answer» `sin^(-1)((x+3)/(4))+c` | |
| 124. |
`intx sqrt(x+x^(2))dx` का मान ज्ञात कीजिए । |
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Answer» `intx sqrt(x+x^(2))dx` यदि `x=A.(d)/(dx)(x+x^(2))+B` `rArr" "x=A(1+2x)+B" …(1)"` अब समीकरण (1 ) के दोनों पक्षों में x की घातों की तुलना करने पर `2A=1 rArr A=(1)/(2)` व`" "A+B=0 rArr B=-(1)/(2)` `rArr" "int xsqrt(x+x^(2))dx=int{(1)/(2)(1+2x)-(1)/(2)}sqrt(x+x^(2))dx` `=(1)/(2)int(1+2x)sqrt(x+x^(2))dx-(1)/(2)intsqrt(x+x^(2))dx` माना `(x+x^(2))= t rArr (1+2x)dx=dt`, तब `=(1)/(2)int sqrtt dt-(1)/(2)int sqrt({(x^(2)+x+(1)/(4))-(1)/(4)})dx` `=(1)/(2).(t^(3//2))/((3//2))-(1)/(2)sqrt({(x+(1)/(2))^(2)-((1)/(2))^(2)})dx` माना `(x+(1)/(2))=u rArr dx =du,` तब , `=(1)/(3)(x+x^(2))^(3//2)-(1)/(2).intsqrt(u^(2)-((1)/(2))^(2))du` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(2){(u)/(2)sqrt(u^(2)-(1)/(4))-(1)/(8)log[u+sqrt(u^(2)-(1)/(4))]}+c` `[" यहाँ "intsqrt(x^(2)-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log[x+sqrt(x^(2)-a^(2))]+c]` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(4)(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))+(1)/(16)log.[(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))]+c` `=(1)/(3)(x+x^(2))^(3//2)-(1)/(8)(2x+1)sqrt(x+x^(2))+(1)/(16)log[(2x+1)/(2).sqrt(x+x^(2))]_c` |
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| 125. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(1+4 cos^(2)x)` |
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Answer» Correct Answer - `(1)/(sqrt5)tan^(-1)((tanx)/(srt5))` `int(1)/(1+4cos^(2)x).dx=int(sec^(2)x)/(sec^(2)x+4).dx` `=int(sec^(2)x)/(5+tan^(2)x)dx` अब माना `" "tanx = t rArr sec^(2)xdx=dt` |
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| 126. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(12-13sinx)` |
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Answer» Correct Answer - `(1)/(5)log[(6 tan x//2-1)/(6 tanx//2 -4)]` `int(dx)/(12-13 sinx)=int(dx)/(12-13 .(2tanx//2)/(1+tan^(2)x//2))` `=int(1+tan^(2)x//2)/(12(1+tan^(2)x//2)(-26tanx//2).dx` `=int(sec^(2)x//2)/(12(1+tan^(2)x//2)-26tanx//2)` अब माना `tanx//2= t rArr sec^(2)x//2 dx=2 dt` |
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| 127. |
`int(dx)/((4+25x^(2)))` |
| Answer» `(1)/(10)tan^(-1)((5x)/(2))+c` | |
| 128. |
`int(dx)/(sqrt(9-25x^(2)))` |
| Answer» `(1)/(5)sin^(-1)((5x)/(3))+c` | |
| 129. |
`x` के सापेक्ष समाकलन करें`(1)/(a cos^(2)x+b sin x cos x +c sin^(2)x)` |
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Answer» फलन `int(dx)/(a cos^(2)x+b sin x cos x+sin^(2)x)` का x के सापेक्ष समाकलन माना `I=int(dx)/(a cos^(2)x+b sin xcos x+c sin^(2)x)` अंश व हर को `cos^(2)x` से भाग देने पर, `I=int(sec^(2)xdx)/(a+b tanx +c tan^(2)x)` माना `tan x = t therefore sec^(2)xdx=dt` `therefore" "I=int(dt)/(a+bt+ct^(2))` यह फलन अब `int(dx)/(ax^(2)+bx+c)` के रूप में परिवर्तित हो गया है, |
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| 130. |
फलन `int(3x+1)/(2x^(2)+x-1)dx` का x के सापेक्ष समाकलन कीजिए। |
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Answer» माना `" "I=int(3x+1)/(2x^(2)+x-1)dx" …(1)"` माना अंश `=A(d)/(dx)("हर")+B` `rArr 3x+1=A(d)/(dx)(2x^(2)+x-1)+B` `=A(4x+1)+B` ltBrgt `3x+1=4Ax+(A+B)` दोनों पक्षों से समान घातीय पदों के गुणांकों की तुलना करने पर, `3=4A rArr A=(3)/(4)` तथा `A+B=1 rArr B=1-A=1-(3)/(4)=(1)/(4)` अतः समीकरण (1 ) से , `I=int(A(d)/(dx)(" हर")+B)/(2x^(2)+x-1)dx` `=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)+(1)/(4)int(dx)/(2x^(2)+x-1)` `=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)dx+(1)/(8)int(dx)/(x^(2)+(1)/(2)x-(1)/(2))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/(x^(2)+(1)/(2)x_(1)/(16)-(1)/(2)-(1)/(16))` `I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/((x+(1)/(4))^(2)-((3)/(4))^(2))` माना `x+(1)/(4)=t" "therefore" "dx=dt" तथा "(3)/(4)=a` `therefore" "I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dt)/(t^(2)-a^(2))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(16a)log((t-a)/(t+1))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(16xx(3)/(4))log((x+(1)/(4)-(3)/(4))/(x+(1)/(4)+(3)/(4)))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(12)log((2x-1)/(2(x+1)))` |
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| 131. |
`(1)/(a^(2)cos^(2)x+b^(2)sin^(2)x)` |
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Answer» फलन `int(dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)` का x के सापेक्ष समाकलन माना `" "I=int(dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)` अंश व हर को `cos^(2)x` से भाग देने पर , `I=int(sec^(2)xdx)/(a^(2)+b^(2)tan^(2)x)=(1)/(a^(2))int(sec^(2)xdx)/(1+((b)/(a)tanx)^(2))` माना `(b)/(a)tanx=t therefore sec^(2)xdx=(a)/(b) dt` `therefore" "I=(1)/(ab)int(dt)/(1+t^(2))=(1)/(ab)tan^(-1)t=(1)/(Ab)tan^(-1)[(b tanx)/(a)]` |
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| 132. |
`int(cosx)/(cos 3x)dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(cosx)/(cos 3x)dx` `=int(cosx)/(4cos^(3)x-3cosx)dx=int(dx)/(4cos^(2)x-3)` अंश व हर को `cos^(2)x` से भाग देने पर `therefore" "I=int(sec^(2)xdx)/(4-3sec^(2)x)` `=int(sec^(2)xdx)/(4-3(1+tan^(2)x))" "` (नोट कीजिए) `=int(sec^(2)xdx)/(1-3tan^(2)x)=int(sec^(2)xdx)/(1-(sqrt3tan x)^(2))` माना `" "sqrt3 tan x = t rArr sec^(2) x dx=(1)/(Sqrt3)dt` `therefore" "I=(1)/(sqrt3)int(dt)/(1-t^(2))=(1)/(sqrt3)int(dt)/((1-t)(1+t))` `=(1)/(2sqrt3)int{(1)/(1-t)+(1)/(1+t)}dt` `=(1)/(2sqrt3)log((1+t)/(1-t))` `=(1)/(2sqrt3)log((1+sqrt3 tan)/(1-sqrt3 tanx))` |
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| 133. |
फलन `int((x+1)dx)/(3-2x-2x^(2))` का x के सापेक्ष समाकलन कीजिए। |
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Answer» माना `" "I=int((x+1)dx)/(3-2x-2x^(2))" …(1)"` माना `" "(x+1)=A(d)/(dx)(" हर ") +B` `=A(d)/(dx)(3-2x-2x^(2))+B` `=A(0-2-4x)+B` या `" "(x+1)=-4Ax+(b-2A)` दोनों पक्षों में समान घातीय पदों की तुलना करने पर, `1=-4A rArr A=-(1)/(4)` तथा `B-2A = 1 rArr B=2A+1=2(-(1)/(4))+1` `therefore" "B=(1)/(2)` अतः समीकरण (1 ) से, `I=int(A(d)/(dx)("हर")+B)/(3-2x-2x^(2))dx` `=-(1)/(4)int((-2-4x)dx)/(3-2x-2x^(2))+(1)/(2)int(dx)/((3-2x-2x^(2)))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/((3)/(2)-(x^(2)+x))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((3)/(2)+(1)/(4))-(x^(2)+x+(1)/(4)))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((sqrt7)/(2))^(2)-(x+(1)/(2))^(2))` माना `(sqrt7)/(2)=a` तथा `x+(1)/(2)=t" "therefore" "dx=dt` `therefore" "I=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dt)/(a^(2)-t^(2))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2a)log((a+t)/(a-t))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2(sqrt7)/(2))log[(sqrt7+2x+1)/(sqrt7-2x-1)]` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4sqrt7)log((sqrt7+2x+1)/(sqrt7-2x-1))` |
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| 134. |
फलन `int(3x+1)/(2x^(2)-2x+3)dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `int(3x+1)/(2x^(2)-2x+3)dx" ...(1)"` अंश को दो फलनों के योग जिसमें प्रथम फलन x का फलन जो कि हर का अवकल गुणांक है तथा दूसरा पद अचर है, के रूप में बदलने के लिये `3x+1=A(d)/(dx)(" हर")+B=A(d)/(dx)(2x^(2)-2x+3)+B` `=A(4x-2)+B=4Ax+(B-2A)` दोनों पक्षों में समान घातीय पदों के गुणकों की तुलना करने पर, `4A=3 rArr A=(3)/(4)` तथा `" "B-2A=1 rArr B=1+2((3)/(4))=(5)/(2)` समीकरण (1 ) में `3x+1=(3)/(4)(4x-2)+(5)/(2)` रखने पर, `I=int((3)/(4)(4x-2)+(5)/(2))/(2x^(2)-2x+3)dx` `=(3)/(4)int((4x-2)dx)/(2x^(2)-2x+3)+(5)/(2)int(dx)/(2x^(2)-2x+3)` `=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)int(dx)/(x^(2)-x+(3)/(2))` `=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)int(dx)/(x^(2)-x+(1)/(4)+(3)/(2)-(1)/(4))` `=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)int(dx)/((x-(1)/(2))^(2)+((sqrt5)/(2))^(2))` माना `x-(1)/(2)=t therefore dx=dt` तथा `(sqrt5)/(2)=a` `I=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)int(dt)/(t^(2)+a^(2))` `=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)(1)/(a)tan^(-1)((t)/(a))` `=(3)/(4)log(2x^(2)-2x+3)+(5)/(4)(2)/(sqrt5)tan^(-1)[(x-(1)/(2))/((sqrt5)/(2))]` `=(3)/(4)log(2x^(2)-2x+3)+(sqrt5)/(2)tan^(-1)((2x-1)/(sqrt5))` |
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| 135. |
`(1)/(a+bsin^(2)x)` |
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Answer» फलन `(1)/(a+b sin^(2)x)` का x के सापेक्ष समाकलन माना `" "I=int(dx)/(a+b sin^(2)x)` अंश व हर को `cos^(2)x` से भाग देने पर, `I=int(sec^(2)xdx)/(a sec^(2)x+b sin^(2)x)` `rArr" "I=int(sec^(2)xdx)/(a+(a+b)tan^(2)x)` `rArr " "I=(1)/(a)int(sec^(2)xdx)/(1+(sqrt((a+b)/(a))tanx)^(2))` माना `sqrt(((a+b)/(a)))tanx=t rArr sec^(2)xdx=sqrt(((a)/(a+b)))dt` `therefore" "I=(1)/(sqrt(a(a+b)))int(dt)/(1+t^(2))=(1)/(sqrt(a(a+b)))tan^(-1)t` `=(1)/(sqrt(a(a+b)))tan^(-1)[sqrt(((a+b)/(a)))tanx]` |
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| 136. |
फलन `int(dx)/(1-x-x^(2))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(1-x-x^(2))` `=int(dx)/(1-(x+x^(2)+(1)/(4)+(1)/(4)))" "` (पूर्ण वर्ग बनाने पर ) `=int(dx)/(1+(1)/(4)-(x+(1)/(2))^(2))=int(dx)/(((sqrt5)/(2))-(x+(1)/(2))^(2))` माना `x+(1)/(2)=t therefore dx=dt" तथा " (sqrt5)/(2)=a` `therefore" "I=int(dt)/(a^(2)-t^(2))` `=(1)/(2a)log((a+t)/(a-t))=(1)/(2(sqrt5)/(2))log[((sqrt5)/(2)+x+(1)/(2))/((sqrt5)/(2)-x-(1)/(2))]` `" "` (x तथा a के मान रखने पर ) `=(1)/(sqrt5)log((sqrt5+2x+1)/(sqrt5-2x-1))` |
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| 137. |
`int((x^(2)+1)dx)/((x+1)^(3)(x-2))` का मान ज्ञात कीजिए। |
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Answer» माना `" "I=int((x^(2)+1)dx)/((x+1)^(3)(x-2))` माना `" "x+1=t rArr x = t-1" तथा " dx=dt` `therefore" "I=int((t-1)^(2)+1)/(t^(3)(t-3))dt=(1)/(t^(3))int(t^(2)-2t+2)/((t-3))dt` अब `t^(2)-2t+2` को `t-3` से भाग दीजिये तथा t की घातों को आरोही क्रम (ascending order ) में रखने पर `=(1)/(t^(3))[-(2)/(3)+(4)/(9)t-(5)/(27)t^(2)+(5t^(3))/(27(t-3))]` `=[-(2)/(3)t^(-3)+(4)/(9)t^(-2)-(5)/(27).(1)/(t)+(5)/(27(t-3))]` `therefore" "int(t^(2)-2t+2)/(t^(3)(t-3))dt=(-2)/(3)intt^(-3)dt+(4)/(9)int t^(-2)dt-(5)/(27)int(dt)/(t)+(5)/(27)int(dt)/((t-3))` `=-(2)/(3)(t^(-2))/(-2)+(4)/(9)(t^(-1))/(-1)-(5)/(27)log t+(5)/(27)log(t-3)` `=(5)/(27)[log(t-3)-logt]+(1)/(3t^(2))-(4)/(9t)` `=(5)/(27)[log(x-2)-log(x+1)]+(1)/(3(x+1)^(2))-(4)/(9(x+1))` `=(5)/(27)log((x-2)/(x+1))+(1)/(3(x+1)^(2))-(4)/(9(x+1))` |
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| 138. |
फलन `int(dx)/(2x^(2)+x-1)` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(2x^(2)+x-1)=(1)/(2)int(dx)/(x^(2)+(1)/(2)x-(1)/(2))` `=(1)/(2)int(dx)/(x^(2)+(1)/(2)x+(1)/(16)-(1)/(16)-(1)/(2))" "` (पूर्ण वर्ग बनाने पर ) `=(1)/(2)int(dx)/((x+(1)/(4))^(2)-((3)/(4))^(2))` माना `x+(1)/(4)=t therefore dx=dt` तथा `(3)/(4)=a` `therefore" "I=(1)/(2)int(dt)/(t^(2)-a^(2))=(1)/(2.2a)log((t-a)/(t+a))` `=(1)/(4xx(3)/(4))log[(x+(1)/(4)-(3)/(4))/(x+(1)/(4)+(3)/(4))]=(1)/(3)log[(2x-1)/(2(x+1))]` |
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| 139. |
फलन `int(dx)/(2sqrt(x(1-x)))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(2sqrt(x(1-x)))=(1)/(2)int(dx)/(sqrt(-(x^(2)-x)))` `=(1)/(2)int (dx)/(sqrt(-(x^(2)-x+(1)/(4)-(1)/(4))))" "` (पूर्ण वर्ग बनाने पर ) `I=(1)/(2)int(dx)/(sqrt(((1)/(2))^(2)-(x-(1)/(2))^(2)))` माना ` x-(1)/(2)=t therefore dx=dt` तथा `(1)/(2)=a` `therefore" "I=(1)/(2)int(dt)/(sqrt(a^(2)-t^(2)))=(1)/(2)sin^(-1)((t)/(a))` `=(1)/(2)sin^(-1)((x-(1)/(2))/((1)/(2)))=(1)/(2)sin^(-1)((2x-1)/(1))` `thereforeI=(1)/(2)sin^(-1)(2x-1)` |
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| 140. |
फलन `int(dx)/(2x^(2)-4x+1)` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(2x^(2)-4x+1)=(1)/(2)int(dx)/(x^(2)-2x+(1)/(2))" "` (नोट कीजिए ) `=(1)/(2)int(dx)/(x^(2)-2x+1-1+(1)/(2))=(1)/(2)int(dx)/((x-1)^(2)-((1)/(sqrt2))^(2))` माना `(x-1)=t therefore dx=dt` तथा `(1)/(sqrt2)=a` `therefore" "I=(1)/(2)int(dt)/(t^(2)-a^(2))` `=(1)/(2)(1)/(2a)log((t-a)/(t+a))=(1)/(4.(1)/(sqrt2))log[((x-1)-(1)/(sqrt2))/((x-1)+(1)/(sqrt2))]` `=(1)/(2sqrt2)log[(sqrt2x-(sqrt2+1))/(sqrt2x-(sqrt2-1))]` |
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| 141. |
फलन `int(dx)/(5+4x+x^(2))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(5+4x+x^(2))=int(dx)/((x^(2)+4x+4)+1)` `=int(dx)/((x+2)^(2)+(1)^(2))` माना `x+2=t rArr dx=dt" तथा " 1=a` `therefore I=int(dt)/(t^(2)+a^(2))=(1)/(a)tan^(-1)((t)/(a))` `=(1)/(1) tan^(-1)((x+2)/(1))" "` (t तथा a के मान रखने पर ) `therefore" "I=tan^(-1)(x+2)` |
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| 142. |
`int(2x+5)/(sqrt(x^(2)+2x+5))dx` का मान ज्ञात कीजिए | |
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Answer» माना `I=int(2x+5)/(sqrt(x^(2)+2x+5))dx` `=int(2x+2)/(sqrt(x^(2)+2x+5))dx+int(3dx)/(sqrt((x^(2)+2x+1)+4))` माना `x^(2)+2x+5= t rArr (2x+2)dx=dt` `therefore I=int(dt)/(sqrtt)+3int(dx)/(sqrt((x+1)^(2)+4))` `=2t^(1//2)+3int(dx)/(sqrt((x+1)^(2)+4))` `=2sqrt(x^(2)+2x+5)+3 int(dx)/(sqrt(4+(x+1)^(2)))` पुनः माना `x+1=2 tan theta rArr dx=2sec^(2)theta d theta` `therefore" "I=2sqrt(x^(2)+2x+5)+3int(2sec^(2)theta d theta)/(sqrt(4+4tan^(2) theta))` `=2sqrt(x^(2)+2x+5)+3 int(sec^(2)theta)/(sec theta)d theta` `=2sqrt(x^(2)+2x+5)+3 log (sec theta+ tan theta)` `=2sqrt(x^(2)+2x+5)+3 log[sqrt(1+((x+1)/(2))^(2))+(x+1)/(2)]` `=2sqrt(x^(2)+2x+5)+3log[(sqrt(x^(2)+2x+5))/(2)+(x+1)/(2)]+c` `=2sqrt(x^(2)+2x+5)+3log[sqrt(x^(2)+2x+5)+(x-1)]+c_(1)` `" "` जहाँ `c_(1)=c-log 2` |
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| 143. |
`(1)/(a+bcos^(2)x)` |
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Answer» फलन `(1)/(a+b cos^(2)x)` का x के सापेक्ष समाकलन माना `I=int(dx)/(a+b cos ^(2)x)` अंश व हर को `cos^(2)x` से भाग देने पर , `I=int(sec^(2)xdx)/(asec^(2)x+b)` `=int(sec^(2)x dx)/((a+b)+atan^(2)x)` `=(1)/((a+b))int(sec^(2)xdx)/(1+(sqrt((a)/((a+b))tanx)^(2)))` माना `sqrt((a)/((a+b)))tanx=t rArr sec^(2)xdx =sqrt(((a+b)/(a)))dt` `therefore" "I=(1)/(sqrt(a(a+b)))int(dt)/(1+t^(2))=(1)/(sqrt(a(a+b)))tan^(-1)t` `=(1)/(sqrt(a(a+b)))tan^(-1)[sqrt((a)/((a+b)))tanx]` |
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| 144. |
फलन `intsqrt((3-2x-2x^(2)))dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=intsqrt((3-2x-2x^(2)))dx` `=intsqrt({3-2(x^(2)+x+(1)/(4)-(1)/(4))})dx" "` (पूर्ण वर्ग बनाने पर ) `=intsqrt(3-2(x+(1)/(2)))dx=intsqrt((7)/(2)-2(x+(1)/(2))^(2))dx` `=sqrt2 int sqrt((sqrt7)/(2)^(2)-(x+(1)/(2))^(2))dx` माना `(sqrt7)/(2)=a` तथा `x+(1)/(2)= t therefore dx=dt` `I=sqrt2 int sqrt((a^(2)-t^(2)))dt` `=sqrt2[(t)/(2)sqrt((a^(2)-t^(2)))+(a^(2))/(2)sin^(-1)((t)/(a))]` `=sqrt2[((x+(1)/(2))/(2))sqrt((7)/(4)-(x+(1)/(2))^(2))+(7)/(8)sin^(_1)((x+(1)/(2))/((sqrt7)/(2)))]` `=(2x+1)/(2sqrt2)sqrt(((7)/(4)-x^(2)-x-(1)/(4)))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` `=(2x+1)/(2sqrt2)[(3)/(2)-x^(2)-x]^((1)/(2))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` `=(2x+1)/(4)[3-2x^(2)-2x]^((1)/(2))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` |
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| 145. |
`int((x-1))/((x+1)(x-2))dx` का मान ज्ञात कीजिए । |
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Answer» `int((x-1))/((x+1)(x-2))dx` आंशिक भिन्नों में व्यक्त करने पर , `((x-1))/((x+1)(x-2))=(A)/((x+1))+(B)/((x-2))` तब,`" "(x-1)=A(x-2)+B(x+1)` यदि `" "x=-1" तब "A=(2)/(3)` और यदि `x = 2" तब "B=(1)/(3)` इस प्रकार `((x-1))/((x+1)(x-2))=(2)/(3(x+1))+(1)/(3(x-2))` `rArr int ((x-1))/((x+1)(x-2))dx=(2)/(3)int(dx)/((x+1))+(1)/(3)int(dx)/((x-2))` `=(2)/(3)log|x+1|+(1)/(3)log|x-2|+c` |
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| 146. |
`int((2x+1)dx)/(sqrt((4x^(2)+8x+6)))` फलन का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=int((2x+1)dx)/(sqrt(4x^(2)+8x+6))" …(1)"` माना `" "2x+1=A(d)/(dx)(4x^(2)+8x+6)+B` `=A(8x+8)+B` `2x+1=8Ax+(8A+B)` दोनों पक्षों में समान घातीय पदों के गुणांकों की तुलना करने पर, `2=8A " "rArr" "A=(1)/(4)` तथा `" "8A+B=1" "rArr" "B=1-8((1)/(4))` `rArr" "B=-1` समीकरण (1 ) से, `I=int(A((d)/(dx))(" हर ")+B)/(sqrt(4x^(2)+8x+6))dx` `=(1)/(4)int(8x+8)/(sqrt(4x^(2)+8x+6))dx-int(dx)/(sqrt(4x+8x+6))` `=(1)/(4)int(D(" हर "))/(sqrt(4x^(2)+8x+6))dx-(1)/(2)int(dx)/(sqrt(x^(2)+2x+(3)/(2)))` `=(1)/(2)sqrt((4x^(2)+8x+6)-(1)/(2)int(dx)/(sqrt((x+1)^(2)+((1)/(sqrt2))^(2)))` `=(1)/(2)sqrt((4x^(2)+8x+6))-(1)/(2)log[(x+1)+sqrt((x+1)^(2)+sqrt((x+1)^(2)+((1)/(sqrt2))^(2)))]` `=(1)/(2)sqrt(4x^(2)+8x+6)-(1)/(2)log[x+1+sqrt(x^(2)+2x+(3)/(2))]` |
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| 147. |
फलन `int(dx)/(sqrt(4x^(2)-x+4))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(sqrt((4x^(2)-x+4)))=(1)/(2)int(dx)/(sqrt((x^(2)-(1)/(4)x+1)))` `=(1)/(2)int(dx)/(sqrt(x^(2)-(1)/(4)x+(1)/(64)+1-(1)/(64)))" "` (पूर्ण वर्ग बनाने पर ) `=(1)/(2)int(dx)/(sqrt((x-(1)/(8))^(2)+((sqrt(63))/(8))^(2))` माना `x-(1)/(8)=t therefore dx=dt` तथा `(sqrt(63))/(8)=a` `therefore" "I=(1)/(2)int(dt)/(sqrt(t^(2)+a^(2)))=(1)/(2)log[t+sqrt(t^(2)+a^(2))]` `=(1)/(2)log[(x-(1)/(8))+sqrt((x-(1)/(8))^(2)+(63)/(64))]` `=(1)/(2)log[(8x-1)/(8)+(1)/(8)sqrt(64x^(2)16x+64)]+c` log 8 को समाकलन स्थिरांक के साथ सम्मिलित करने पर , `=(1)/(2)log[(8x-1)+sqrt(64x^(2)-16x+64)]+c` |
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| 148. |
फलन `int(dx)/(sqrt(2+x-3x^(2)))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(sqrt(2+x-3x^(2)))=int(dx)/(sqrt(2-3(x^(2)-(x)/(2)))` `=int(dx)/(sqrt(2-3{x^(2)-(x)/(3)+(1)/(36)-(1)/(36)})" "` (पूर्ण वर्ग बनाने पर ) `=int(dx)/(sqrt(2+(1)/(12)-3(x-(1)/(6))^(2))=(1)/(sqrt3)int(dx)/(sqrt(((5)/(6))^(2)-(x-(1)/(6))^(2)))` माना `x-(1)/(6)=t therefore dx = dt` तथा `(5)/(6)=a` `therefore" "I=(1)/(sqrt3)int(dt)/(sqrt(a^(2)-t^(2)))=(1)/(sqrt3)sin^(-1)((t)/(a))` `=(1)/(sqrt3)sin^(-1){(x-(1)/(6))/((5)/(6))}=(1)/(sqrt3)sin^(-1)[(6x-1)/(5)]` |
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| 149. |
`int(dx)/(a^(2)+x^(2))` का मान ज्ञात कीजिए | |
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Answer» माना `I=int(dx)/(a^(2)+x^(2))=(1)/(a^(2))int(dx)/(1+((x)/(a))^(2))` माना `(x)/(a)=tan theta` `therefore" "dx=a sec^(2) theta d theta` `therefore" "I=(a)/(a^(2))int(sec^(2) theta d theta)/(1+tan^(2) theta)=(1)/(a) int d theta =(1)/(a) theta` अतः `int(dx)/(a^(2)+x^(2))=(1)/(a) tan^(-1)((x)/(a))` `" "[because tan theta =(x)/(a)rArr theta = tan^(-1)((x)/(a))]` |
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| 150. |
फलन `int(2x+1)/(sqrt((2x^(2)+x-3)))dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=int(2x+1)/(sqrt(2x^(2)+x-3))dx" ...(1)"` माना `2x+1=A(d)/(dx)(2x^(2)+x-3)+B` `=A(4x+1)+B` `=4Ax+(A+B)` दोनों पक्षों में समान घातीय पदों के गुणांकों की तुलना करने पर, `2=4A rArr A=(1)/(2)` तथा `" "A+B=1" " rArr" "B=(1)/(2)` अतः समीकरण (1 ) से, `I=(1)/(2)int(4x+1)/(sqrt(2x^(2)+x-3))dx +(1)/(2)int(dx)/(sqrt(2x^(2)+x-3))` `=(1)/(2)((2x^(2)+x-3)^(-(1)/(2)+1))/(-(1)/(2)+1)+(1)/(2sqrt2)int(dx)/(sqrt(x^(2)+(1)/(2)x-(3)/(2)))` `=sqrt(2x^(2)+x-3)+(1)/(2sqrt2)int(dx)/(sqrt((x^(2)+(1)/(2)x+(1)/(16)-(1)/(16)-(3)/(2))))` `" "` (पूर्ण वर्ग बनाने पर) `=sqrt((2x^(2)+x-3))+(1)/(2sqrt2)int(dx)/(sqrt((x+(1)/(4))^(2)-((5)/(4))^(2))` माना `x+(1)/(4)= t rArr dx = dt` तथा `(5)/(4)=a` `therefore" "I=sqrt(2x^(2)+x-3)+(1)/(2sqrt2)int(dt)/(sqrt(t^(2)-a^(2)))` `=sqrt((2x^(2)+x-3))+(1)/(2sqrt2)log[t+sqrt(t^(2)-a^(2))]` `=sqrt((2x^(2)+x-3))+(1)/(2sqrt2)log[(x+(1)/(4))+sqrt((x+(1)/(4))^(2)-((5)/(4))^(2))]` `=sqrt((2x^(2)+x-3))+(1)/(2sqrt2)log[(4x+1)+sqrt(16x^(2)+8x-24)]` `" "` (log 4 को अचर में सम्मिलित करने पर ) |
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