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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.
| 51. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(x-1)sqrt(x^(2)+x+1)` |
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Answer» Correct Answer - `(1)/(3)(x^(2)+x+1)^(3//2)-(3)/(2)[((2x+1)/(4))sqrt((x^(2)+x+1))+(3)/(8)log{(x+(1)/(2))+sqrt((x^(2)+x+1))}]` `int(x-1)sqrt(x^(2)+x+1).dx` `=int[(1)/(2)(2x+1)-(3)/(2)]sqrt(x^(2)+x+1).dx` `=(1)/(2)int(2x+1)sqrt(x^(2)+x+1).dx -(3)/(2)intsqrt(x^(2)+x+1).dx` |
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| 52. |
`(x^(2)+1)/(x^(4)+x^(2(+1)` |
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Answer» Correct Answer - `(1)/(sqrt3)tan^(-1)((x^(2)-1)/(xsqrt3))` `int(x^(2)+1)/(x^(4)+x^(2)+1)dx=int(1+(1)/(x^(2)))/([x-(1)/(x)]^(2)+3)dx` अब माना `x-(1)/(x)=t " "rArr" "1+(1)/(x^(2))dx=dt` |
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| 53. |
सिद्ध कीजिए कि (i) `int(1)/(1+cos^(2)x)dx=(1)/(sqrt2)tan(-1)((tanx)/(sqrt2))+c` |
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Answer» (i) `int(dx)/(sqrt3 sin x+cosx)dx` `=(1)/(2)int(1)/(((sqrt3)/(2)sinx+(1)/(2)cosx))dx=(1)/(2)int(dx)/((sin.(pi)/(3)sinx+cos.(pi)/(3)cosx))` `=(1)/(2)int(dx)/(cos(x-(pi)/(3)))=(1)/(2)int sec(x-(pi)/(3))dx` `=(1)/(2)log tan ((x-(pi)/(3))/(2)+(pi)/(4))+c` |
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| 54. |
`int(x^(2)+1)/((x-1)^(3)(x+3))dx` |
| Answer» `(3)/(8)log|x-1|-(1)/(2(x-1))+(5)/(8)log|x+3|+c` | |
| 55. |
`int(1-x^(2))/(x(1-2x))dx` |
| Answer» `(x)/(2)+log|x|-(3)/(4)log|1-2x|+c` | |
| 56. |
`int(dx)/((x+1)^(2)(x^(2)+1))` |
| Answer» `(1)/(2)log|x+1|-(1)/(2(x+1))-(1)/(4)log|x^(2)+1|+c` | |
| 57. |
फलन `int(dx)/(5+4 cos x)` का x के सापेक्ष समाकलन कीजिए। |
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Answer» माना `I=int(dx)/(5+4cosx)=int(dx)/(5+4((1-tan^(2)x//2)/(1+tan^(2)x//2)))` `=int((1+tan^(2)x//2)dx)/(5(1+tan^(2)x//2)+4(1-tan^(2)x//2))` `=int(sec^(2)x//2dx)/(9+tan^(2)x//2)=(1)/(9)int(sec^(2)x//2 dx)/(1+((tanx//2)/(3))^(2))` माना `(tanx//2)/(3)= t therefore sec^(2)x//2 dx=6dt` `therefore I=(6)/(9)int(dt)/(1+t^(2))=(2)/(3)tan^(-1)((tanx//2)/(3))` |
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| 58. |
`int(x/((x-1)^(2)(x+2)))`dx |
| Answer» `(2)/(9)log|(x-1)/(x+2)|-(1)/(3(x-1))+c` | |
| 59. |
फलन `int(dx)/((asin x+b cosx)^(2))` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=int(dx)/((a sin x +b cosx)^(2))` `=int(dx)/(cos^(2)x(a tanx+b)^(2))=int(sec^(2)xdx)/((atanx+b)^(2))` माना `a tan x+b = t therefore sec^(2)xdx=(1)/(a)dt` `thereofore " "I=(1)/(a)int(dt)/((t)^(2))=(1)/(a)int t^(-2)dt` `=-(1)/(a(t))=(-1)/(a(atanx+b))` |
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| 60. |
`int(2x)/((x^(2)+1)(x^(2)+3))dx` का मान ज्ञात कीजिए । |
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Answer» माना `x^(2)=t rArr 2xdx=dt` `therefore" "int(2x)/((x^(2)+1)x^(2)+3))dx=int(1)/((t+1)(t+3))dt` ltBrgt `=(1)/(2)int[(1)/((t+1))-(1)/((t+3))]dt` `=(1)/(2)int(1)/((t+1))dt-(1)/(2)int(1)/((t+3))dt` `=(1)/(2)log(t+1)-(1)/(2)log(t+3)+c` `=(1)/(2)log((t+1)/(t+3))+c=(1)/(2)log((x^(2)+1)/(x^(2)+3))+c` |
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| 61. |
फलन `int(x^(2))/((x^(2)+1)(3x^(2)+1))dx` का x के सापेक्ष समाकलन |
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Answer» यदि अंश व हर दोनों में x की केवल समघात ही हो तब `x^(2)=t` रखकर फलन का आंशिक भिन्नो में वियोजन कीजिए। अतः `x^(2)=t` रखने पर `(x^(2))/((x^(2)+1)(3x^(2)+1))=(t)/((t+1)(3t+1))` `rArr" "(t)/((t+1)(3t+1))=(A)/((t+1))+(B)/((3t+1))` `therefore" "t=A(3t+1)+B(t+1)" ...(1)"` समीकरण (1) में `t+1=0 rArr 1=-1` रखने पर `-a=A(-3+1)" "rArr" "A=(1)/(2)` तथा `3t+1=0 rArr t=-(1)/(3)` रखने पर `-(1)/(3)=B(-(1)/(3)+1)" "rArr -(1)/(3)=(2)/(3)B` `therefore" "B=-(1)/(2)` `therefore" "(t)/((t+1)(3t+1))=(1)/(2(t+1))-(1)/(2).(1)/((3t+1))` `therefore int(x^(2)dx)/((x^(2)+1)(3x^(2)+1))=(1)/(2)int(dx)/(1+x^(2))-(1)/(2)int(dx)/(3x^(2)+1)` `=(1)/(2)int(dx)/(1+x^(2))-(1)/(6)int(dx)/(x^(2)+((1)/(sqrt3))^(2))` `=(1)/(2)tan^(-1)(x)-(1)/(2sqrt3)tan^(-1)(x sqrt3)` |
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| 62. |
फलन `int(dx)/(5+4 sin x)` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `I=int(dx)/(5+4 sinx)` `=int(dx)/(5(sin^(2)x//2+cos^(2)x//2)+8sin x//2 cos x//2)` अंश हर को `cos^(2)x//2` से भाग देने पर , `I=int(sec^(2)x//2dx)/(5(1+tan^(2)x//2)+8tanx//2)` `=(1)/(5)int(sec^(2)x//2dx)/(1+tan^(2)x//2+(8)/(5)tanx//2)` माना `tan x//2 = t therefore sec^(2)x//2 dx=2dt` `therefore" "I=(2)/(5)int(dt)/(1+t^(2)+(8)/(5)t)=(2)/(5)int(dt)/(1+(t^(2)+(8)/(5)t+(16)/(25)-(16)/(25)))` `" "` (पूर्ण वर्ग बनाने पर ) `=(2)/(5)int(dt)/((1-(16)/(25))+(t+(4)/(5))^(2))=(2)/(5)int(dt)/(((3)/(5))^(2)+(t+(4)/(5))^(2))` माना `t+(4)/(5)=u" "rArr" "dt=du" तथा " (3)/(5)=a` `therefore" "I=(2)/(5)int(du)/(a^(2)+u^(2))=(2)/(5a)tan^(-1)((u)/(a))` `=(2)/(5xx(3)/(5))tan^(-1)[(5t+4)/(3)]=(2)/(3)tan^(-1)[(5tan x//2 +4)/(3)]` द्वितीय विधि - `I=int(dx)/(5+4sinx)=int(dx)/(5+(4xx2tanx//2)/(1+tan^(2)x//2))` `=int((1+tan^(2)x//2)dx)/(5(1+tan^(2)x//2)+8 tan x//2)` `=(1)/(5)int(sec^(2)x//2dx)/(1+tan^(2)x//2+(8)/(5)tanx//2)` `=(1)/(5)int(sec^(2)x//2dx)/(1+(tan^(2)x//2+(8)/(5)tanx//2+(16)/(25)-(16)/(25)))` `" "` (पूर्ण वर्ग बनाने पर ) `=(1)/(5)int(sec^(2)x//2dx)/(((3)/(5))^(2)+(tanx//2+(4)/(5))^(2))` माना `" "(3)/(5)=a` तथा `tan x//2+(4)/(5)=u" "therefore" "sec^(2).(x)/(2)dx=2du` `therefore" "I=(2)/(5)int(du)/(a^(2)+u^(2))=(2)/(5a)tan^(-1)((u)/(a))` `=(2)/(3)tan^(-1)[(5tanx//2+4)/(3)]` |
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| 63. |
फलन `int(cosx)/(cos 3x)` का मान ज्ञात कीजिए। |
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Answer» `int(cosx)/(cos 3x)dx=int(cosx)/(4cos^(3)x-3cosx)dx=int(1)/(4cos^(2)x-3)dx` `=int(1)/(4cos^(2)x-3(cos^(2)x+sin^(2)x))dx` `=int(1)/(cos^(2)x-3sin^(2)x)dx=int(1)/(cos^(2)x(1-3tan^(2)x))dx` `=int(sec^(2)x)/(1-3tan^(2)x)dx=(1)/(sqrt3)int(dt)/(1-t^(2))` ( यहाँ `sqrt3 tanx =t rArr sec^(2)x dx=(1)/(sqrt3)dt`) `=(1)/(sqrt3).(1)/(2.1)log((1+t)/(1-t))+c` `=(1)/(2sqrt3)log((1+sqrt3 tanx)/(1-sqrt3 tanx))+c` `=(1)/(2sqrt3)log((1+sqrt3tanx)/(1-sqrt3tanx))+c` |
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| 64. |
`int(sinx)/((1+cos x)(2+3 cos x))` का मान ज्ञात कीजिए । |
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Answer» माना `cos x = t rArr" "-sin x dx =dt` `therefore" "int(sinx)/((1+cos x)(2+3cosx))dx=-int(dt)/((1+t)(2+3t))` `=-int[(-1)/((1+t))+(3)/((2+3t))]dt` `=int(1)/((1+t))dt-3int(1)/((2+3t))dt` `=log (1+t)-3.(1)/(3)log(2+3t)+c` `=log((1+t))/((2+3t))+c=log ((1+cos x)/(2+3 cos x))+c` |
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| 65. |
`int(x)/((x^(4)-x^(2)+1))dx` का मान ज्ञात कीजिए । |
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Answer» `int(x)/((x^(4)-x^(2)+1))dx` यदि `t=x^(2) rArr 2x dx = dt,` तब `int(x)/((x^(4)-x^(2)+1))dx=(1)/(2).int(dt)/((t^(2)-t+1))` `=(1)/(2).int(dt)/({(t-(1)/(2))^(2)+((sqrt3)/(2))^(2)})` `=(1)/(2).(1)/(((sqrt3)/(2)))tan^(-1).((t-(1)/(2)))/(((sqrt3)/(2)))+c` `=(1)/(sqrt3)tan^(-1)((2t-1)/(sqrt3))+c` `=(1)/(sqrt3)tan^(-1)((2x^(2)-1)/(sqrt3))+c` |
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| 66. |
`int(1)/(3+sin 2x)dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(1)/(3+sin 2x)dx` `=int(1)/(3+2sin x cosx)dx" "(because sin 2x = 2 sin x cos x)` `=int(dx)/(3(cos^(2)x+sin^(2)x)+2sin x cos x)` `" "[because 3 = 3.1=3(cos^(2)x +sin^(2)x)]` `=int(sec^(2)x)/(3tan^(2)x+2 tan x+3)dx` (अंश व हर को `cos^(2)x` से भाग देने पर)) माना `tan x = t rArr sec^(2)xdx=dt` `therefore" "I=int(1)/(3t^(2)+2t+3)dt=(1)/(3)int(1)/(t^(2)+(2)/(3)t+1)dt` `=(1)/(3)int(1)/((t+(1)/(3))^(2)+((2sqrt2)/(3))^(2))` `=(1)/(3)int(1)/(t^(2)+(2)/(3)t+1)dt` `=(1)/(3)int(1)/((t+(1)/(3))^(2)+((2sqrt2)/(3))^(2))` `=(1)/(3).(1)/(((2sqrt2))/(3))tan^(-1)((t+(1)/(3))/((2sqrt2)/(3)))+c` `=(1)/(2sqrt2)tan^(-1)((3t+1)/(2sqrt2))+c` `=(1)/(2sqrt2)tan^(-1)((3 tan x+1)/(2sqrt2))+c` |
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| 67. |
`int(1)/(x^(2)+x+1)` का मान ज्ञात कीजिए । |
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Answer» हम जानते हैं कि `x^(2)+x+1=x^(2)+x+(1)/(4)+(1-(1)/(4))=(x+(1)/(2))^(2)+(x+(1)/(2))^(2)+((sqrt3)/(2))^(2)` `therefore (1)/(x^(2)+x+1)dx=int(1)/((x+(1)/(2))^(2)+((sqrt3)/(2))^(2))dx` `=int(1)/(t^(2)+((sqrt3)/(2))^(2))dt" "` जहाँ `x+(1)/(2)=t rArr dx=dt` `=(1)/(sqrt3//2)tn^(-1)((t)/(sqrt3//2))+c` `=(2)/(sqrt3)tan^(-1)[(2)/(sqrt3)(x+(1)/(2))]+c` `=(2)/(sqrt3)tam^(-1)((2x+1)/(sqrt3))+c` |
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| 68. |
`int(dx)/(sqrt(15-8x^(2)))` का मान ज्ञात कीजिए । |
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Answer» `int(dx)/(sqrt(15-8x^(2)))=(1)/(sqrt8)int(dx)/(sqrt((15)/(8)-x^(2)))` `=(1)/(2sqrt2)sin^(-1){(x)/(sqrt((15)/(8)))}+c=(1)/(2sqrt2)sin^(-1)(sqrt((8)/(15))x)+c` |
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| 69. |
`int(1)/(x[6 (logx)^(2)+7log x +2])dx` का मान ज्ञात कीजिए । |
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Answer» माना `log x = t rArr (1)/(x)dx=dt` `therefore" "int(1)/(x[6(log x)^(2)+7log x+2])dx=int(1)/(6t^(2)+7t+2)dt` `=(1)/(6)int(1)/(t^(2)+(7)/(6)t+(1)/(3))dt` `=(1)/(6)int(1)/((t+(7)/(12))^(2)+(1)/(3)-(49)/(144))fy` `=(1)/(6)int(1)/((t+(7)/(12))^(2)-((1)/(12))^(2))dt` `=(1)/(6).(1)/(2.(1)/(12))log[(t+(7)/(12)-(1)/(12))/(t+(7)/(12)+(1)/(12))]+c_(1)` `=log((t+1//2)/(t+2//3))+c_(1)` `=log((3(2t+1))/(2(3t+2)))+c_(1)` `=log((2logx+1)/(3log x+2))+log (3)/(2)+c_(1)` `=log((2log x+1)/(3logx+2))+c" जहाँ "c=c_(1)+log(3)/(2)` |
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| 70. |
`int(cos x)/(sqrt(sin^(2)x-2 sin x-3))dx` का मान ज्ञात कीजिए । |
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Answer» `int(cosx)/(sqrt(sin^(2)x-2sin x-3))dx` यदि `sin x = t rArr cos x dx =dt` तब , `=int(dt)/(sqrt(t^(2)-02t-3))=int(dt)/(sqrt((t1)^(2)-2^(2))` `=log[(t-1)+sqrt((t-1)^(2)-2^(2))]+c` `=log[(sinx-1)+sqrt(sin^(2)x-2 sin x-3)]+c` |
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| 71. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(2x+1)sqrt(x^(2)-x+1)` |
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Answer» Correct Answer - `(2)/(3)(x^(2)-x+1)^(3//2)+((2x-1)/(2))sqrt((x^(2)-x+1))+(3)/(4)log[(2x-1)/(2)+sqrt((x^(2)-x+1))]` `int(2x+1)sqrt(x^(2)-x+1).dx` `=int[(2x-1)+2]sqrt(x^(2)-x+1).dx` `=int(2x-1)sqrt(x^(2)-x+1).dx+2 intsqrt(x^(2)-x+1).dx` |
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| 72. |
`(x+1)/(sqrt(x^(2)+1))` |
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Answer» Correct Answer - `sqrt((x^(2)+1))+log[x+sqrt((x^(2)+1))]` `int(x+1)/(sqrt(x^(2)+1)).dx=(1)/(2)int(2x)/(sqrt(x^(2)+1))dx+int(dx)/(sqrt(x^(2)+1^(2)))` |
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| 73. |
`int(1-x^(2))/(1+x^(4))dx` |
| Answer» `(1)/(2sqrt2)log|(sqrt2.x+x^(2)+1)/(sqrt2. x-x^(2)-1)|+c` | |
| 74. |
`int (5x+3)/(sqrt(x^(2)+4x+10))dx` |
| Answer» `5sqrt(x^(2)+x+1)+2log|x+(1)/(2)+sqrt(x^(2)+x+1)|+c` | |
| 75. |
`int(dx)/(sqrt(2x^(2)+4x+6))` |
| Answer» `(1)/(sqrt2)log|(x+1)+sqrt(x^(2)+2x+3)|+c` | |
| 76. |
`intxsqrt(x^(4)+1)dx` का मान ज्ञात कीजिए । |
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Answer» `intxsqrt(X^(4)+1)dx` यदि ` x^(2)=t rArr xdx=(dt)/(2),` तब, `intxsqrt(x^(4)+1)dx=(1)/(2)intsqrt(t^(2)+1)dt` `=(1)/(2)[(t)/(2)sqrt(t^(2)+1)+(1)/(2)log[t+sqrt(t^(2)+1)]+c` `=(x^(2))/(4)sqrt(x^(4)+1)+(1)/(4)log[x^(2)+sqrt(x^(4)+1)]+c` |
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| 77. |
`int(sqrt(tanx)+sqrt(cotx))dx` का मान ज्ञात कीजिए । |
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Answer» `int(sqrt(tanx+sqrt(cotx)))dx` `=int(sqrt(tanx)+(1)/(sqrt(tanx)))dx=int((tanx+1)/(sqrt(tanx)))dx` माना `x=t^(2) rArr x = tan^(-1) t^(2)` व `dx=(2t)/((1+t^(4)))dt` `=int((t^(2)+1))/(t).(2t)/((1+t^(4)))dt=2int((t^(2)+1))/((t^(2)+1))dt` `=2int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt=2int((1+(1)/(t^(2))))/((t-(1)/(t))^(2)+2)dt` माना `(t-(1)/(t))=u rArr (1+(1)/(t^(2)))dt=du,` तब `2int((1+(t)/(t^(2))))/((t-(1)/(t))^(2)+2)dt=2int(du)/((u^(2)+2))` `=2.(1)/(sqrt2)tan^(-1).(u)/(sqrt2)+c` `=sqrt2 tan^(-1).((t-(1)/(t)))/(sqrt2)+c` `=sqrt2 tan^(-1).((t^(2)-1))/((sqrt2t))+c` `=sqrt2 tan^(-1)((tanx-1)/(sqrt2 tanx))+c` |
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| 78. |
`(1)/(x(x^(n)+1))` |
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Answer» Correct Answer - `(1)/(n)log[(x^(n))/(1+x^(n))]` `int(dx)/(x(x^(n)+1))=int(x^(n-1))/(x x^(n-1)(x^(n)+1))dx=int(x^(n-1))/(x^(n)(x^(n)+1))dx` अब `x^(n)=t` रखने पर |
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| 79. |
`int(sec^(2)x)/(sqrt(16+tan^(2)x))dx` |
| Answer» `log|tanx +sqrt(tan^(2)x+16)|+c` | |
| 80. |
`int(dx)/(sqrt(8-4x-2x^(2)))` |
| Answer» `(1)/(sqrt2)sin^(-1)((x+1)/(sqrt5))+c` | |
| 81. |
`int(sin 2x)/((1- cos 2x)(2- cos 2x))dx` |
| Answer» `(1)/(2)log|1-cos2x|-(1)/(2)log|2-cos2x|+c` | |
| 82. |
`int(dx)/(2+sin^(2)x)` |
| Answer» `(1)/(sqrt6)tan^(-1)((sqrt3. tanx)/(sqrt2))+c` | |
| 83. |
फलनों का समाकलन कीजिए - `(1)/(1+x+x^(2)+x^(3))` |
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Answer» Correct Answer - `(1)/(2)log(1+x)-(1)/(4)log(1+x^(2))+(1)/(2)tan^(-1)x` `(1)/(1+x+x^(2)+x^(3))=(1)/(2(1+x))-(x)/(2(1+x^(2)))+(2)/(2(1+x^(2)))` |
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| 84. |
फलनों का समाकलन कीजिए - `(4x-3)/(3x^(2)+2x-5)` |
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Answer» Correct Answer - `(2)/(3)log(3x^(2)+2x-5)-(13)/(24)log((3x-3)/(3x+5))` `2x^(2)+2x+1=((2x+1)^(2))/(2)+(1)/(2)` |
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| 85. |
फलनों का समाकलन कीजिए - `(1)/(1+3e^(x)+2e^(2x))` |
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Answer» Correct Answer - `log((2e^(x)+1)/(e^(x)+1))` माना `e^(x)=t` तब `(1)/(1+3e^(x)+2e^(2x))=(1)/(1+3t+2t^(2))` `=(1)/((2t+1)(t+1))=(2)/((2t+1)(t+1))=(2)/(2t+1)-(1)/((t+1))` |
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| 86. |
फलनों का समाकलन कीजिए - `(sec^(2)x)/((3+tanx)(4+tanx))` |
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Answer» Correct Answer - `log((3+tanx)/(4+tanx))` माना `tan x= t rArr sec^(2)x dx = dt` तब `int(sec^(2)xdx)/((3+tanx)(4+tanx))=int(dt)/((3+t)(4+t))=int[(1)/(3+t)-(1)/(4+t)]dt` |
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| 87. |
फलनों का समाकलन कीजिए - `(1)/(2x^(2)+3x-4)` |
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Answer» Correct Answer - `(1)/(sqrt(41))log[(4x+3-sqrt(41))/(4x+3+sqrt(41))]` `int(1)/(2x^(2)+3x-4)dx=8int(dx)/((4x+3)^(2)-(sqrt(41))^(2))` |
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| 88. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((x^(2)+3x+1)))` |
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Answer» Correct Answer - `log(2x+3)+2sqrt(x^(2)+3x+1)` `x^(2)+3x+1=x^(2)+3x+(9)/(4)+1-(9)/(4)=(x^(2)+3x+(9)/(4))-(5)/(4)` `=(x+(3)/(2))^(2)-(5)/(4)=(1)/(4)[(2x+3)^(2)-(sqrt5)^(2)]` |
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| 89. |
फलनों का समाकलन कीजिए - `(1)/(x^(4)-1)` |
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Answer» Correct Answer - `(1)/(4)log((x-1)/(x+1))-(1)/(2)tan^(-1)x` `(1)/(x^(4)-1)=(1)/((x^(2)-1)(x^(2)+1))=(1)/((t-1)(t+1))" "(" जब " x^(2)=t)` `=(1)/(2(t-1))-(1)/(2(t+1))=(1)/(2(x^(2)=1))-(1)/(2(x^(2)+1))` अब समाकलन कीजिए । |
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| 90. |
फलनों का समाकलन कीजिए - `(x^(2))/(x^(6)+x^(3)-2)` |
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Answer» Correct Answer - `(1)/(9)log((x^(3)-1)/(x^(3)+2))` माना `x^(3)=t rArr 3x^(2)dx=dt` तब `int(x^(2))/(x^(6)+x^(3)-2)dx=(1)/(3)int(dt)/(t^(2)+t-2)=(1)/(3)int(dt)/((t-1)(t+2))` `=(1)/(3)int[(1)/(t-1)-(1)/(t+2)]dt` |
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| 91. |
फलनों का समाकलन कीजिए - `(x^(2)+1)/(x^(2)-1)` |
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Answer» Correct Answer - `x+log((x-1)/(x+1))` `(x^(2)+1)/(x^(2)-1)=int((x^(2)-1)+2)/((x^(2)-1))dx=int(1+(2)/(x^(2)-1))dx=int1dx+2int(1)/(x^(2)-1)dx` `=x+2xx(1)/(2)log((x-1)/(x+1))=x+log((x-1)/(x+1))` |
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| 92. |
फलनों का समाकलन कीजिए - `(1)/(x^(2)+x+1)` |
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Answer» Correct Answer - `(2)/(sqrt3)tan^(-1)((2x+1)/(sqrt3))` `int(1)/(x^(2)+x+1)dx=int(dx)/((x+(1)/(2))^(2)((sqrt3)/(2))^(2))=(1)/(sqrt3)tan^(-1)((x+(1)/(2))/((sqrt3)/(2)))` |
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| 93. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((2x^(2)+2x+3)))` |
| Answer» `(1)/(sqrt2)[log(2x+1)+sqrt2 sqrt(2x^(2)+2x+3)]` | |
| 94. |
फलनों का समाकलन कीजिए - `(1)/(2x^(2)-8x+25)` |
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Answer» Correct Answer - `(1)/(sqrt(34))tan^(-1)(sqrt2(x-2))/(sqrt(17))` `2x^(2)-8x+25=2(x^(2)-4x+(25)/(2))=2[(x=2)^(2)+(17)/(2)]` |
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| 95. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((x^(2)+2x+2)))` |
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Answer» Correct Answer - `log[(x+1)+sqrt((x^(2)+2x+2))]` `int(1)/(sqrt((x^(2)+2x+2)))dx=int(1dx)/(sqrt((x+1)^(2)+1))=int(dt)/(sqrt(t^(2)+1))` `"जब " t=x+t rArr dx=dt` |
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| 96. |
फलनों का समाकलन कीजिए - `(1)/(x^(2)+2x+5)` |
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Answer» Correct Answer - `(1)/(2)tan^(-1)((x+1)/(2))` `int(dx)/(x^(2)+2x+5)=int(dx)/((x+1)^(2)+2^(2))=(1)/(2)tan^(-1)((x+1)/(2))+c` |
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| 97. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(1)/(sqrt((5x-x^(2)-6)))` |
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Answer» Correct Answer - `sin^(-1)(2x-5)` `5x-x^(2)-6=-x^(2)+5x-6=-(x^(2)-5x+6)` `=-(x^(2)-5x+(25)/(4)-(25)/(4)+6)=(1)/(4){(1-(2x-5)^(2))}` |
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| 98. |
फलनों का x के सापेक्ष समाकलन कीजिए - `sqrt(2x^(2)-5x-1)` |
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Answer» Correct Answer - `(4x-5)/(8)sqrt(2x^(2)-5x-1)-(33)/(16sqrt2)log[(4x-5)/(4)+(sqrt(2x^(2)-5x-1))/(sqrt2)]` `2x^(2)-5x-1=2(x^(2)-(5)/(2)x-(1)/(2))=2[(x-(5)/(4))^(2)-((33)/(16))]` |
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| 99. |
फलनों का x के सापेक्ष समाकलन कीजिए - `(2x+3)/(sqrt(x^(2)+4x+1))` |
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Answer» Correct Answer - `2sqrt((x^(2)+4x+1))-log[(x+2)+sqrt((x^(2)+4x+1))]` `int(2x+3)/(sqrt(x^(2)+4x+1))dx=int(2x+4)/(sqrt(x^(2)+4x+1))dx-int(dx)/(sqrt(x^(2)+4x+1))` |
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| 100. |
फलनों का समाकलन कीजिए - `(1)/(2x^(2)+x-1)` |
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Answer» Correct Answer - `(1)/(3)log((2x-1)/(x+1))` समीकरण `(1)/(2x^(2)+x-1)` को आंशिक भिन्नों में व्यक्त करने पर `(1)/(2x^(2)+x-1)=(3)/(2(2x-1))-(1)/(3(x+1))` अब समाकलन कीजिए |
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