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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.
| 151. |
फलन `int(dx)/(sqrt(1-4x-2x^(2)))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(sqrt(1-4x-2x^(2)))` `=int(dx)/(sqrt(1-2(x^(2)+2x+1-1))" "` (नोट कीजिए ) `=int(dx)/(sqrt(3-2(x+1)^(2)))=(1)/(sqrt2)int(dx)/(sqrt((sqrt((3)/(2)))^(2)-(x+1)^(2)))` माना `" "x+1=t rArr dx=dt` तथा `" "sqrt((3)/(2))=a` `therefore" "I=(1)/(sqrt2)int(dt)/(sqrt(a^(2)-t^(2)))=(1)/(sqrt2)sin^(-1)((t)/(a))` `=(1)/(sqrt2)sin^(-1)[(sqrt2(x+1))/(sqrt3)]` |
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| 152. |
फलन `intsqrt((5+4x-x^(2))dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=intsqrt((5+4x-x^(2)))dx` `=intsqrt(5-(x^(2)-4x+4-4))dx" "` (पूर्ण वर्ग बनाने पर ) `=intsqrt(9-(x-2)^(2))dx=intsqrt((3)^(2)-(x-2)^(2))dx` माना `3=a` तथा `x-2=t therefore dx=dt` `therefore I=intsqrt((a^(2)-t^(2)))dx=(t)/(2)sqrt((a^(2)-t^(2)))+(1)/(2)sqrt(a^(2)-t^(2))+(1)/(2)a^(2)sin^(-1)((t)/(a))` `=((x-2)/(2))sqrt(9-(x-2)^(2))+(1)/(2)(3)^(2)sin^(-1)((x-2)/(3))` `=(1)/(2)(x-2)sqrt(5+4x-x^(2))+(9)/(2)sin^(-1)((x-2)/(3))` |
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| 153. |
`int((3x-2))/((x+1)^(2)(x+3))dx` का मान ज्ञात कीजिए । |
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Answer» `int((3x-2))/((x+1)^(2)(x+3))dx` आंशिक भिन्नों में व्यक्त करने पर `((3x-2))/((x+1)^(2)(x+3))=(A)/((x+1))+(B)/((x+1)^(2))+(C)/((x+3))" ...(1)"` यदि `" "x=-3" तब "C=-(11)/(4)` यदि `" "x=-1" "तब "B=-(5)/(2)` अब, समीकरण (1 ) के दोनों, पक्षों में x की घातों की तुलना करने पर `A+C=0, A=(11)/(4),` तब `((3x-2))/((x+1)^(2)(x+3))=(11)/(4(x+1))-(5)/(2(x+1)^(2))-(11)/(4(x+3))` `rArr" "int((3x-2))/((x+1)^(2)(x+3))dx=(11)/(4)int(dx)/((x+1))-(5)/(2)int(1)/((x+1)^(2))dx-(11)/(4)int(dx)/((x+3))` `=(11)/(4)log(x+1)+(5)/(2(x+1))-(11)/(4)log(x+3)+c` `=(11)/(4).log((x+1)/(x+3))+(5)/(2(x+1))+c` |
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| 154. |
फलन `int(2x-5)sqrt((2+3x-x^(2)))dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=int(2x-5)sqrt((2+3x-x^(2)))dx` माना `" "(2x-5)=A(d)/(dx)(2+3x-x^(2))+B` `=A(3-2x)+B` `2x-5=-2Ax+(3A+B)` दोनों पक्षों में समान घातीय पदों के गुणांकों की तुलना करने पर `-2A=2` `therefore" "A=-1` तथा `" "3A+B=-5` `rArr" "B=-3A-5` `therefore" "B=-2` `therefore I=int[-(3x-2x)sqrt((2x+3x-x^(2)))-2sqrt(2+3x-x^(2))]dx` `=-int(3-2x)sqrt((2+3x-x^(2)))dx-2intsqrt((2+3x-x^(2)))dx` `=-int(3-2x)sqrt((2+3x-x^(2)))dx-2intsqrt(2-(x^(2)-3x+(9)/(4)-(9)/(4)))dx` `" "` (पूर्ण वर्ग बनाने पर ) `=-((2+3x-x^(2))^((1)/(2)+1))/((1)/(2)+1)-2intsqrt((17)/(4)-(x-(3)/(2))^(2))dx` `=-(2)/(3)(2+3x-x^(2))^((3)/(2))-2intsqrt(((sqrt(17))/(2))^(2)-(x-(3)/(2))^(2))dx` माना `(sqrt(17))/(2)=a` तथा `x-(3)/(2)=t therefore dx=dt` `therefore" "I=-(2)/(3)(2+3x-x^(2))^(3//2)-2intsqrt((a^(2)-t^(2)))dt` `=-(2)/(3)(2+3x-x^(2))^(3//2)-2[(t)/(2)sqrt(a^(2)-t^(2))+(a^(2))/(2)sin^(-1)((t)/(a))]` `=-(2)/(3)(2+3x-x^(2))^(3//2)+(3-2x)/(2)sqrt(2+3x-x^(2))-(17)/(4)sin^(-1)(2x-3)/(sqrt(17))` |
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| 155. |
`int(x^(2))/((x^(2)+4)(x^(2)+9))dx` का मान ज्ञात कीजिए । |
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Answer» `int(x^(2))/((x^(2)+4)(x^(2)+9))dx` माना `x^(2)=y` तथा आंशिक भिन्नों में व्यक्त करने पर ltbgt `(y)/((y+4)(y+9))=(A)/((y+4))+(B)/((y+9))" …(1)"` `y=A(y+9)+B(y+4)` यदि `" "y=-4," तब "A=-(4)/(5)` और यदि `" "y=-9," तब "B=(9)/(5)` अब समीकरण (1 ) से `(y)/((y+4)(y+9))=(-4)/(5(y+4))+(9)/(5(y+9))` `rArr" "(x^(2))/((x^(2)+4)(x^(2)+9))=(-4)/(5(x^(2)+4))+(9)/(5(x^(2)+9))` `rArr int (x^(2))/((x^(2)+4)(x^(2)+9))dx=-(4)/(5)int(1)/((x^(2)+4))dx+(9)/(5)int(1)/((x^(2)+9))dx` `=-(4)/(5)tan^(-1).(x)/(2).(1)/(2)+(9)/(5)tan^(-1).(x)/(3).(1)/(3)+c` `=-(2)/(5)tan^(-1).(x)/(2)+(3)/(5)tan^(-1).(x)/(3)+c` |
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| 156. |
`int(x^(2))/(sqrt(x^(6)+a^(6)))dx` का मान होगा -A. `(1)/(3)log|x^(3)+sqrt(x^(6)+a^(6))|+c`B. `(2)/(3)log|x^(3)+sqrt(x^(6)+a^(6))|+c`C. `(5)/(3)log|x^(2)+sqrt(x^(6)-a^(6))|+c`D. `(1)/(3)log|x^(3)-sqrt(x^(6)-a^(6))|+c` |
| Answer» Correct Answer - A | |
| 157. |
`int (dx)/(a^(2)-x^(2))` का मान ज्ञात कीजिए | |
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Answer» माना `I=int(dx)/(a^(2)-x^(2))=int(dx)/((a-x)(a+x))` आंशिक भिन्नों में वियोजन `(1)/((a-x)(a+x))=(A)/((a-x))+(B)/((a+x))` `rArr" "1=A(a+x)+B(a-x)" ...(1)"` माना समीकरण (1 ) के दोनों पक्षों में `a-x=0 rArr x = a` रखने पर,, `1=A(2a)+0 rArr A=(1)/(2a)` पुनः समीकरण (2 ) के दोनों पक्षों में `a+x=0 rArr x = -a` रखने पर, `1=0 +B(2a)" "therefore" "B=(1)/(2a)` अतः `(1)/((a-x)(a+x))=(1)/(2a(a-x))+(1)/(2a(a+x))` `therefore" "int(dx)/((a-x)(a+x))=(1)/(2a)int(dx)/((a-x))+(1)/(2a)int(dx)/((a+x))` `=-(1)/(2a)log(a-x)+(1)/(2a)log(a+x)` `=(1)/(2a)log((a+x)/(a-x))` `(because log M-log N=log(M)/(N))` `therefore" "int(dx)/(a^(2)-x^(2))=(1)/(2a)log((a+x)/(a-x)),` यदि `a gt x` |
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| 158. |
`int(1)/(3x^(2)+13x-10)dx` का मान होगा -A. `(1)/*(15)log|(3x+2)/(x+5)|+c`B. `(1)/(17),pg|(3x-2)/(x-5)|+c`C. `(1)/(17)log|(3x-2)/(x+5)|+c`D. `(1)/(17)log|(3x+2)/(x-5)|+c` |
| Answer» Correct Answer - C | |
| 159. |
`in(dx)/(x^(2)-a^(2))` का मान ज्ञात कीजिए यदि `x gt a` |
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Answer» माना `I=int_(dx)/(x^(2)-a^(2))=int(dx)/((x-a)(x+a))" …(1)"` आंशिक भिन्नों में वियोजन माना `" "(1)/((x-a)(x+a))=(A)/((x-a))+(B)/((x+a))` `rArr" "1=A(x+a)+B(x-a)" ...(2)"` समीकरण (2 ) के दोनों पक्षों में `x-a=0 rArr x=a` रखने पर, `1=A(a+a)+0` `rArr" "A=(1)/(2a)` पुनः समीकरण (2 ) के दोनों पक्षों में `x+a=0 rArr x=-a` रखने पर, `1=0+B(-2a)` `therefore" "B=-(1)/(2a)` अतः `(1)/((x-a)(x+a))=(1)/(2a)[(1)/((x-a))-(1)/((x+a))]` `therefore" "int(dx)/((x-a)(x+a))=(1)/(2a) int[(1)/((x-a))-(1)/((x+a))]dx` `=(1)/(2a)[int(dx)/((x-a))-int(dx)/((x+a))]` `=(1)/(2a)[log(x-a)-log(x+a)]` `=(1)/(2a)log((x-a)/(x+a))` `(because log M log N = log(M)/(N))` `therefore" "int(dx)/(x^(2)-a^(2))=(1)/(2a)log((x-a)/(x+a))`, यदि `xgt a` |
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| 160. |
`int(1)/(sqrt(7-6x-x^(2)))dx` का मान होगा -A. `cos^(-1)((x+3)/(4))+c`B. `(1)/(2)sin^(-1)((x+3)/(4))+c`C. `tan^(-1)((x+3)/(5))+c`D. `sin^(-1)((x+3)/(4))+c` |
| Answer» Correct Answer - D | |
| 161. |
`int((tan theta +tan^(3) theta))/((1+ tan^(3) theta))d theta ` का मान ज्ञात कीजिए । |
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Answer» `int((tan theta +tan^(3) theta))/((1+ tan^(3) theta))d theta ` अब `(tan theta +tan^(3) theta)/(1+tan^(3) theta)=(tan theta(1+ tan^(2) theta))/((1+tan^(3) theta))=(tan theta sec^(2) theta)/((1+ tan^(3) theta))` तब `int((tan theta+ tan^(3) theta))/((1+tan^(3) theta))d theta= int(tan theta sec^(2) theta)/( (1+tan^(3) theta))d theta" ...(1)"` यदि `tan theta = t rArr sec^(2) theta d theta = dt` अब समीकरण (1 ) से `=int(t)/((1+t^(3)))dt=int(t)/((1+t)(1-t+t^(2)))dt` आंशिक भिन्नो में व्यक्त करने पर `(t)/((1+t)(1-t+t^(2)))=(A)/((1+t))+((Bt+C))/((1-t+t^(2)))` `t=A(1-t+t^(2))+(Bt+c)(1+t)" ...(2)"` यदि `" "t=-1" तब "A=-(1)/(3)` व अब समीकरण (2 ) के दोनों पक्षों में t की घातों की तुलना करने पर `A+B=0 rArr B=(1)/(3)` व `A+C=0 rArr C=(1)/(3)` अब `(t)/((1+t)(1-t+t^(2)))=(-1)/(3(1+t))+(((1)/(3)t+(1)/(3)))/((1-t+t^(2)))` `int(t)/((1+t)(1-t+t^(2)))dt` `=-(1)/(3)int(dt)/((1+t))+(1)/(6)int(2t)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))` `=-(1)/(3)int(dt)/((1+t))+(1)/(6)int((2t-1)+1)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))` `=-(1)/(3)int(dt)/((1+t))+(1)/(6)int((2t-1))/((t^(2)-t+1))dt+(1)/(2)int(dt)/((t^(2)-t+1))` `=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2)int(dt)/((t^(2)-t+(1)/(4))+(3)/(4))` `=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2)int(dt)/((t-(1)/(2))^(2)+((sqrt3)/(2))^(2))` `=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2).(2)/(sqrt3)tan^(-1).((t-(1)/(2)))/(((sqrt3)/(2)))+c` `=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(sqrt3)tan^(-1)((2t-1)/(sqrt3))+c` `=-(1)/(3)log(1+tan theta)+(1)/(6)log (tan^(2)theta - tan theta+1)+(1)/(sqrt3)tan^(-1)((2 tan theta-1)/(sqrt3))+c` |
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| 162. |
फलन `int(dx)/(1+3sin^(2)x)` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=int(dx)/(1+3sin^(2)x)` अंश व हर को `cos^(2)x` से भाग देने पर ,, `I=int(sec^(2)xdx)/(sec^()x+3tan^(2)x)=int(sec^(2)x)/(1+4tan^(2)x)dx` माना `tanx = t therefore sec^(2)x dx = dt (because sec^(2)x=1+tan^(2)x)` `therefore" "I=int(dt)/(1+4t^(2))=(1)/(2)tan^(-1)(2t)` `=(1)/(2)tan^(-1)(2 tanx)+c` |
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| 163. |
`int(1)/(sqrt((x-a)(x-b)))dx` का मान होगा -A. `log|x+(a+b)/(2)+sqrt((x-a)(x-b))|+c`B. `log|x-(a+b)/(2)+sqrt((x-a)(x-b))|+c`C. `log|x+(a+b)/(2)-sqrt((x-a)(x-b))|+c`D. `log|x-(a+b)/(2)-sqrt((x-a)(x-b))|+c` |
| Answer» Correct Answer - B | |
| 164. |
`int((x^(2)+1))/((x+1)^(2))dx` का मान ज्ञात कीजिए । |
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Answer» `int((x^(2)+1))/((x+1)^(2))dx=int((x^(2)+1))/((x^(2)+2x+1))` `=int{1-(2x)/((x+1)^(2))}dx` अब, आंशिक भिन्नों में व्यक्त करने पर , `(2x)/((x+1)^(2))=(A)/((x+1))+(B)/((x+1)^(2))` `rArr" "2x=A(x+1)+B` x की घातों की तुलना करने पर `A= 2 " तथा " B=-2` इस प्रकार `(2x)/((x+1)^(2))=(2)/((x+1))-(2)/((x+1)^(2))` `int{1-(2x)/((x+1)^(2))}dx=int{1-(2)/((x+1))+(2)/((x+1)^(2))}dx` `=x-2 log(x+1)-(2)/((x+1))+c` |
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| 165. |
`intcos^(-3//7)x sin^(-11//7)xdx` का मान होगा -A. `log|sin^(4//7)x|+c`B. `(4)/(7)tan^(4//7)x+c`C. `-(7)/(4)tan^(-4//7)x+c`D. `log|cos^(3//7)x|+c` |
| Answer» Correct Answer - C | |
| 166. |
`int(sqrtx)/(sqrt(a^(3)-x^(3)))dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(sqrtx)/(sqrt(a^(3)-x^(3)))dx=int(sqrtx)/(sqrt((a^(3//2))-(x^(3//2))))dx` `x^(3//2)=a^(3//2)t` रखने पर `rArr" "(3)/(2)x^(1//2)dx=a^(3//2)dt=sqrtx dx =(2)/(3)a^(3//2)dt` `therefore" "I=int((2//3)a^(3//2))/(sqrt((a^(3//2))^(2)-(a^(3//2)t)^(2)))` `=(2)/(3)a^(3//2)int(dt)/(a^(3//2)sqrt(1-t^(2)))` `=(2)/(3)int(dt)/(sqrt(1-t^(2)))=(2)/(3)sin^(-1)((t)/(1)+c)` `=(2)/(3)sin^(-1)((x^(3//2))/(a^(3//2)))+c" "[t=(x^(3//2))/(a^(3//2))" रखने पर "]` `=(2)/(3)sin^(-1)(sqrt((x^(3))/(a^(3))))+c` |
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| 167. |
`int(dx)/(sinx+sin2x)` का मान ज्ञात कीजिए । |
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Answer» यदि `" " I=int(dx)/(sinx+sin2x)` `=int(dx)/(sinx+2 sinx cosx)` `=int(dx)/(sinx(1+2 cosx))` माना `cos x = t rArr - sin x dx = dt` तब `" "I=int(-(dt)/(sinx))/(sinx(1+2t))=-int(dt)/(sin^(2)x(1+2t))` `=-int(dt)/((1-t^(2))(1+2t))" ...(1)"` समीकरण `(1)/((1-t^(2))(1+2t))` को आंशिक भिन्नों में व्यक्त करने पर , `(1)/((1-t^(2))(1+2t))=(A)/(1-t)+(B)/(1+t)+(C)/(1+2t)` `rArr 1 = A(1+t)(1+2t)+B(1-t)(1+2t)+c(1-t)(1+t)" ...(2)"` समीकरण (2 ) में `t=1, -1` व `-(1)/(2)` रखने पर `A=(1)/(6), B=-(1)/(2),C=(4)/(3)` तब समीकरण (1 ) व (2 ) से `I=-int[(1)/(6).(1)/(1-t)-(1)/(2).(1)/(1+t)+(4)/(3).(1)/(1+2t)]dt` `=-[-(1)/(6)log(1-t)-(1)/(2)log|1+t|+(4)/(3).(1)/(2)log|1+2t|+c]` `=(1)/(6)log|1-cos x|+(1)/(2)log|1+cosx|-(2)/(3)log|1+2cos x|+c` |
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| 168. |
`int(dx)/((sinx - sin 2x))` का मान ज्ञात कीजिए । |
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Answer» `int(dx)/((sinx - sin 2x))= int(dx)/((sinx-2sin xcosx))` `=int(dx)/(sinx(1-2cosx))=int(sinx)/(sin^(2)x(1-2 cosx))dx` `=int(sinx)/((1-cos^(2)x)(1-2 cosx))dx` यदि `cos x = t ` तथा `-sin x dx=dt,` तब `=-int(dt)/((1-t^(2))(1-2t))=int(dt)/((t-1)(t+1)(1-2t))" ...(1)"` अब, आंशिक भिन्नो में व्यक्त करने पर `(1)/((t-1)(t+1)(1-2t))=(A)/((t-1))+(B)/((t+1))+(C)/((1-2t))" ...(2)"` `1=A(t+1)(1-2t)+B(t-1)(1-2t)+C(t-1)(t+1)` यदि `t=1," तब "A=-(1)/(2)` यदि `t=-1," तब "B=-(1)/(6)` और यदि `t=(1)/(2)," तब "C=-(4)/(3)` अब, समीकरण (1 ) व (2 ) से, `int(dt)/((t-1)(t+1)(1-2t))=-(1)/(2)int(dt)/((t-1))-(1)/(6)int(dt)/((t+1))` `=-(1)/(2)log(t-1)-(1)/(6)log(t+1)+(2)/(3)log(1-2t)+c` `=-(1)/(2)log(cosx-1)-(1)/(6)log(cosx+1)+(2)/(3)log(1- 2 cos x)+c` |
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| 169. |
`int(x+3)sqrt(3-4x-x^(2)).dx` का मान ज्ञात कीजिए । |
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Answer» यदि `" "I=int(x+3)srt(3-4x-x^(2))dx` अब `" "(d)/(dx)(3-4x-x^(2))=-4 - 2x` अतः `" "x+3 =-(1)/(2)(-4 -2x)+1` `therefore I=intsqrt(3-4x-x^(2)).{-(1)/(2)(-4-2x)+1}dx` `=-(1)/(2)intsqrt(3-4x-x^(2))(-4 - 2x)dx+intsqrt(3-4x-x^(2))dx` `=-(1)/(2)I_(1)+I_(2)` यहाँ `I_(1)=intsqrt(3-4x-x^(2)).(-4-2x)dx` `3-4x-x^(2)=t rArr -4-2xdx=dt` `therefore" "I_(1)=intt^(1//2)dt=(t^(3//2))/(3//2)=(2)/(3)(3-4x-x^(2))^(3//2)" ...(2)"` तथा `" "I_(2)=intsqrt(3-4x-x^(2))dx` `=intsqrt(7-(x+2)^(2))dx` `=intsqrt((sqrt7)^(2)-(x+2)^(2))dx` `=((x+2)sqrt(3-4x-x^(2)))/(2)+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c" ...(3)"` अब, समीकरण (1 ), (2 ) व (3 ) से `I=-(1)/(3)(3-4x-x^(2))^(3//2)+(1)/(2)(x+2)sqrt(3-4x-x^(2))+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c` |
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| 170. |
`int(xdx)/(x^(2)+4x+5)` का मान होगा -A. `(1)/(2)log|(x^(2)+4x+5)|+2tan^(-1)x+c`B. `(1)/(2)log|(x^(2)+4x+5)|-tan^(-1)(x+2)+c`C. `(1)/(2)log|(x^(2)+4x+5)|+tan^(-1)(x+2)+c`D. `(1)/(2)log|(x^(2)+4x+5)|-2tan^(-1)(x+2)+c` |
| Answer» Correct Answer - D | |
| 171. |
`int(3x)/((1+2x^(4)))dx` |
| Answer» `(3)/(2sqrt2)tan^(-1)(sqrt2. x^(2))+c` | |
| 172. |
`int(2x-5)sqrt(x^(2)-4x+3)dx` |
| Answer» `(2)/(3)(x^(2)-4x+3)^(3//2)-(1)/(2)(x-2)sqrt(x^(2)-4x+3)+(1)/(2)log|x-2+sqrt(x^(2)-4x+3)|+c` | |
| 173. |
`(x)/((x-1)(x-2)(x-3))` |
| Answer» `(1)/(2)log|x-1|-2log|x-2|+(3)/(2)log|x-3|+c` | |
| 174. |
`int(dx)/(x^(2)+16)` |
| Answer» `(1)/(4)tan^(-1).(x)/(4)+c` | |
| 175. |
`int(x^(2))/(sqrt(x^(6)+a^(6)))` |
| Answer» `(2)/(3)log|x^(3)+sqrt(x^(6)+a^(6))|+c` | |
| 176. |
`int(x^(2))/((1-x^(6)))` |
| Answer» `(1)/(6)log|(1+x^(3))/(1-x^(3))|+c` | |
| 177. |
`int(x)/((x^(4)-9))dx` |
| Answer» `(1)/(12)log|(x^(2)-3)/(x^(2)+3)|+c` | |
| 178. |
`int(3x+1)/((x+2)(x-2)^(2))dx` |
| Answer» `-(5)/(16)log(x+2)+(5)/(16)log|x-2|-(7)/(4(x-2))+c` | |
| 179. |
`int(x^(2)+x+1)/((x+2)(x^(2)+1))dx` |
| Answer» `(3)/(5)log|x+2|+(1)/(5)log|x^(2)+1|+(1)/(5)tan^(-1)x+c` | |
| 180. |
`int(2 cos x)/(3sin^(2)x)dx` |
| Answer» `-(2)/(3)"cosec x"+c` | |
| 181. |
`(1)/(x^(2)-9)` |
| Answer» `(1)/(6)log|(x-3)/(x+3)|+c` | |
| 182. |
`(2x)/(x^(2)+3x+2)` |
| Answer» `4log|x+2|-2log|x+1|+c` | |
| 183. |
`int(2x^(2)+1)/(x^(2)(x^(2)+4))dx` |
| Answer» `-(1)/(4x)+(7)/(8)tan^(-1).(x)/(2)+c` | |
| 184. |
`int(x+2)/(sqrt(x^(2)+5x+6))dx` |
| Answer» `sqrt(x^(2)+5x+6)-(1)/(2)log|(x+(5)/(2))+sqrt(x^(2)+5x+6)|+c` | |
| 185. |
`int(sin 2x)/((1+sinx)(2+sinx))dx` |
| Answer» `-2log|1+sinx|+4log|2+sinx|+c` | |
| 186. |
`int(x^(2))/((x^(2)+4)(x^(2)+25))dx` |
| Answer» `-(1)/(14)tan^(-1).(x)/(2)+(8)/(35)tan^(-1)((x)/(5))+c` | |
| 187. |
`int(2x+1)/(4-3x-x^(2))dx` |
| Answer» `-(3)/(5)log|1-x|-(7)/(5)log|4-x|+c` | |
| 188. |
`int((5x-2))/((1+2x+3x^(2)))dx` का मान ज्ञात कीजिए । |
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Answer» `int((5x-2))/((1+2x+3x^(2)))dx` अब माना कि `(5x-1)=A.(d)/(dx)(1+2x+3x^(2))+B` तब `" "(5x-2)=A(6x+2)+B" …(1)"` अब, समीकरण (1 ) के दोनों पक्षों के x की घातों की तुलना करने पर `6A=5" तथा "2A+B=-2` `rArr" "A=(5)/(6)" तथा "B=(-11)/(3)` तब `" "int ({(5)/(6)(6x+2)-(11)/(3)})/((1+2x+3x^(2)))dx` `=(5)/(6)int((6x+2))/((1+2x+3x^(2)))dx-(11)/(3)int(dx)/((3x^(2)+2x+1))` `=(5)/(6)log(1+2x+3x^(2))-(11)/(3).(1)/(3)int(dx)/((x^(2)+(2)/(3)x+(1)/(3)))` `=(5)/(6)log(1+2x+3x^(2))-(11)/(9)int(dx)/({(x+(1)/(3))^(2)+((1)/(3)-(1)/(9))})` `=(5)/(6)log(1+2x+3x^(2))-(11)/(9)int(dx)/({(x+(1)/(3))^(2)+((sqrt2)/(3))^(2)})+c` `=(5)/(6)log(1+2x+3x^(2))-(11)/(9).(1)/(((sqrt2)/(3)))tan^(-1){(x+(1)/(3))/((sqrt2)/(3))}+c` `=(5)/(6)log(1+2x+3x^(2))-(11)/(3sqrt2)tan^(-1)((3x+1)/(sqrt2))+c` |
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| 189. |
`int(x^(2)+x+1)/((x+2)(x+1)^(2))dx` |
| Answer» `3log|x+2|-2log|x+1|-(1)/((x+1))+c` | |
| 190. |
फलनों का समाकलन कीजिए - `(2x-1)/(2x^(2)+2x+1)` |
| Answer» Correct Answer - `(1)/(2)log(2x^(2)+2x+1)-2tan^(-1)(2x+1)` | |
| 191. |
`int(x+3)sqrt(3-4x-x^(2))dx` का मान ज्ञात कीजिए । |
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Answer» `int(x+3)sqrt(3-4x-x^(2))dx` अब माना कि `x(x+3)=A.(d)/(dx)(3-4x-x^(2))+B` `rArr" "(x+3)=A(-4-2x)+B" …(1)"` अब समीकरण (1 ) के दोनों पक्षों में x की घातों की तुलना करने पर `-2A=1 rArr A=-(1)/(2)` तथा `-4A+B=3 rArr B=1` अब `int(x+3)sqrt(3-4x-x^(2))dx` `=int{-(1)/(2)(-4-2x)+1}.sqrt(3-4x-x^(2))dx` `=-(1)/(2)int(-4-2x)sqrt(3-4x-x^(2))dx+intsqrt(3-4x-x^(2))dx` यदि `3-4x-x^(2)=t" व "dt=(-4-2x)dx` `=-(1)/(2)intsqrtt dt +int sqrt(7-(4x+x^(2)+4))dx` `=(-(1)/(2)xxt^(3//2)xx(2)/(3))+intsqrt((sqrt7)^(2)-(x+2)^(2))dx` `=-(1)/(3)(3-4x-x^(2))^(3//2)+(1)/(2)(x+2)sqrt(3-4x-x^(2))+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c` `[" यहाँ "intsqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(A^(2)-x^(2))+(a^(2))/(2)sin^(-1).(x)/(a)+c]` |
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| 192. |
`int((2x+1))/((4-3x-x^(2)))dx` का मान ज्ञात कीजिए । |
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Answer» `int((2x+1))/((4-3x-x^(2)))dx` अब माना कि `(2x+1)=A.(d)/(dx)(4-3x-x^(2))+B` `rArr" "(2x+1)=A(-3-2x)+B" …(1)"` अब समीकरण (1 ) के दोनों पक्षों के x की घातों की तुलना करने पर `-2A=2, -3A+B=1` तब `" "A=-1" तथा "B=-2` अब `" "int((2x+1))/((4-3x-x^(2)))dx=int{((-1).(-3-2x)-2)/((4-3x-x^(2))}dx` `=-int((-3--2x))/((4-3x-x^(2)))dx-2 int(dx)/((4-3x-x^(2)))` `=-log (4-3x-x^(2))+2 int(dx)/((x^(2)+3x-4))` `=-log(4-3x-x^(2))+2 int(dx)/({(x+(3)/(2))^(2)-((5)/(2))^(2)})` `=-log (4-3x-x^(2))+(2)/(5)log(((x+(3)/(2))-(5)/(2))/((x+(3)/(2))+(5)/(2)))+c` `=-log (4-3x-x^(2))+(2)/(5)log((x-1)/(x+4))+c` |
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| 193. |
फलनों का समाकलन कीजिए - `(x+1)/(x^(2)+4x+5)` |
| Answer» Correct Answer - `(1)/(2)log(x^(2)+4x+5)-tan^(-1)` | |
| 194. |
`(x^(2)+x+1)/(x^(2)-1)` |
| Answer» `(x^(2))/(2)+(1)/(2)log|x+1|+(3)/(2)log|x-1|+c` | |
| 195. |
फलनों का समाकलन कीजिए - `(cosx)/((1+sinx)(2+sinx))` |
| Answer» Correct Answer - `log((1+sinx)/(2+sinx))` | |
| 196. |
फलनों का समाकलन कीजिए - `(3x)/((x-1)(x-2)(x-3))` |
| Answer» Correct Answer - `(3)/(2)log(x-1)-6log(x-2)+(9)/(2)log(x-3)` | |
| 197. |
फलनों का समाकलन कीजिए - `(2x-3)/(x^(2)+3x-18)` |
| Answer» Correct Answer - `(1)/(3)log(x-3)+(5)/(3)log(x+6)` | |
| 198. |
`(3x-1)/((x+2)^(2))` |
| Answer» `3log|x+2|+(7)/(x+2)+c` | |
| 199. |
`(1)/(x^(4)-1)` |
| Answer» `(1)/(4)log|(x-1)/x+1|-(1)/(2)tan^(-1)x+c` | |
| 200. |
फलनों का समाकलन कीजिए - `(x^(2))/(x^(2)-1)(x^(2)+2)` |
| Answer» Correct Answer - `(1)/(6)log((x-1)/(x+1))+(2)/(3sqrt2)tan^(-1)((x)/(sqrt2))` | |