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InterviewSolution
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.
| 301. |
`int " log"_(e) " x dx "` |
| Answer» Correct Answer - `x (log_(e) x-1)+c` | |
| 302. |
`int(sinx)/(1+sin x) dx` |
| Answer» Correct Answer - `sec x-tan x+x+c` | |
| 303. |
`intsqrt(1-cos 2x) dx` |
| Answer» Correct Answer - `-sqrt(2) cos x+c` | |
| 304. |
`inttan^(- 1)((3x-x^3)/(1-3x^2)) dx` |
| Answer» Correct Answer - `3x tan^(-1) x-(3)/(2) log (1+x^(2)) +c` | |
| 305. |
`int(2x^(2)+3)/((x^(2)-1)(x^(2)+4))dx=alog((x+1)/(x-1))+b"tan"^(-1)(x)/(2)` , then (a,b) isA. `(1, -1)`B. `(-1, 1)`C. `((1)/(2),-(1)/(2))`D. `((1)/(2),(1)/(2))` |
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Answer» Correct Answer - D Let `l=int(2x^(2)+3)/((x^(2)-1)(x^(2)+4))dx=int(dx)/(x^(2)-1)+int(dx)/(x^(2)+4)` `" "[because (2x^(2)+3)/((x^(2)-1)(x^(2)+4))=(1)/(x^(2)-1)+(1)/(x^(2)+4)]` `rArr" "l=(1)/(2)log((x-1)/(x+1))+(1)/(2)tan^(-1).(x)/(2)+C` `But" "l=a log((x-1)/(x+1))+b tan^(-1)((x)/(2))+C` `therefore" "a=(1)/(2), b=(1)/(2)` |
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| 306. |
Evaluate: `int(x+1)/(x^2+4x+5) dx` |
| Answer» Correct Answer - `(1)/(2) log |3x^(2)+4x+5 |-tan^(-1) (x+2)+c` | |
| 307. |
` (i) int(1)/((1+x^(2))tan ^(-1) x )dx " "(ii) int(e^(tan^(-1)x))/(1+x^(2))dx` |
| Answer» Correct Answer - `(i) log | tan^(-1)x|+c" "(ii) e^(tan^(-1)x)+c` | |
| 308. |
Evaluate : `int x^n log x dx.` |
| Answer» Correct Answer - ` (x^(n+1))/(n+1)[log x-(1)/(n+1)]+c` | |
| 309. |
`inte^(x+3) dx` |
| Answer» Correct Answer - `e^(x+3)+c.` | |
| 310. |
`int(e^(x) (1+x))/(sin^(2) (x e^(x)))dx` |
| Answer» Correct Answer - `-cot (x e^(x)) +c` | |
| 311. |
`int(2-3 cos x)/(sin^(2) x)dx` |
| Answer» Correct Answer - `-2 cot x+3 cosec x+c`. | |
| 312. |
The value of `int(3x+2)/((x-2)^(2)(x-3))dx` is equal toA. `11log (x-3)/(x-2)-(8)/(x-2)+C`B. `11log (x+3)/(x+2)-(8)/(x-2)+C`C. `11log(x-3)/(x-2)+(8)/(x-2)+C`D. `11log (x+3)/(x+2)+(8)/(x-2)+C` |
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Answer» Correct Answer - C Let l`=int(3x+2)/((x-2)^(2)(x-3))dx" …(i)"` Again, let `(3x+2)/((x-2)^(2)(x-3))=(A)/((x-2))+(B)/((x-2)^(2))+(C)/((x-3))" …(ii)"` `rArr" "3x+2=A(x-2)(x-3)+B(x-3)+C(x-2)^(2)" ...(iii)"` On putting the values of `x=2,3` respectively, we get `3xx2+2=B(2-3)` `rArr B=-8` and `3xx3+2=C(3-2)^(2)` `rArr" "C=11` On equation the coefficient of `x^(2)` in Eq. (iii), we get `0=A+C" "rArr" "A=-11` On putting the values of A, B and C in Eq. (ii), we get `(3x+2)/((x-2)^(2)(x-3))=-(11)/((x-2))-(8)/((x-2)^(2))+(11)/((x-3))` `therefore" "l=int[-(11)/((x-2))-(8)/((x-2)^(2))+(11)/(x-3)]dx` `-11log(x-2)+(8)/((x-2))+11log(x-3)+C` `11log((x-3)/(x-2))+(8)/((x-2))+C` |
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| 313. |
`inte^(sin theta)[log sin theta+"cosec"^(2)theta]cos theta d theta` is equal toA. `inte^(sin theta)[log sin theta+"cosec"^(2)theta]+C`B. `e^(sin theta)[log sin theta+"cosec "theta]+C`C. `e^(sin theta)[log sin theta-"cosec "theta]+C`D. `e^(sin theta)[log sin theta - "cosec"^(2) theta]+C` |
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Answer» Correct Answer - C Let `l=inte^(sin theta)(log sin theta)cos theta d theta+int e^(sin theta)"cosec"^(2)thetacos theta d theta` Put `sin theta = t rArr cos theta d theta = dt` `therefore" "l=int e^(t) log t dt +int e^(t)t^(-2)dt` `=logt e^(t)-int(e^(t))/(t)dt+(e^(t)t^(-1))/(-1)-int(e^(t)t^(-))/(-1)dt` `=e^(t)(logt-(1)/(t))+c` `=e^(sin theta)(log sin theta - "cosec" theta )+C` |
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| 314. |
`int (2x-sin 2x)/(1-cos 2x) dx` |
| Answer» Correct Answer - `-x cot x+c` | |
| 315. |
`int(3x+1)/(2x^(2)+x-1)dx` |
| Answer» Correct Answer - `(3)/(4) log|2x^(2) +x-1|+(1)/(15) log |(2x-1)/(2x+2)|+c` | |
| 316. |
`int(sec^(2)x)/(sqrt(tanx))dx` |
| Answer» Correct Answer - `2sqrt(tanx)+c` | |
| 317. |
`int(1+cos 2x)/(1-cos 2x)dx` |
| Answer» Correct Answer - `-cot x-x+c` | |
| 318. |
`int((x+1)(2x-3))/(x) dx` |
| Answer» Correct Answer - `x^(2) -x-3log x+c` | |
| 319. |
If `int(1)/(x^(3)+x^(4))dx=(A)/(x^(2))+(B)/(x)+log|(x)/(x+1)|+C` , thenA. `A=(1)/(2), B=1`B. `A=1, B=-(1)/(2)`C. `A=-(1)/(2), B=1`D. A = 1, B = 1 |
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Answer» Correct Answer - C `int(dx)/(x^(4)+x^(3))=int((x+1)-x)/(x^(3)(x+1))dx=int((1)/(x^(3))-(1)/(x^(2)(x+1)))dx` `=int((1)/(x^(3))-(1)/(x^(2))+(1)/(x(x+1)))dx` `=int((1)/(x^(3))-(1)/(x^(2))+(1)/(x)-(1)/(x+1))dx` `=-(1)/(2x^(2))+(1)/(x)+log|x|-log|x+1|+C` `-(1)/(2x^(2))+(1)/(x)+log|(x)/(x+1)|+C` `therefore" "A=-(1)/(2) and B=1` |
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| 320. |
`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+C`, thenA. `f(x)=log(logx),g(x)=(1)/(logx)`B. `f(x)=logx, g(x)=(1)/(logx)`C. `f(x)=(1)/(logx),g(x)=log(logx)`D. `f(x)=(1)/(xlogx),g(x)=(1)/(logx)` |
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Answer» Correct Answer - A Given, `int[log(logx)+(1)/((logx)^(2))]dx` `=x[f(x)-g(x)]+C` `LHS=int underset("II")(1).log underset("I")((logx))dx+int(1)/((logx)^(2))dx` On integration by parts, we get `xlog(logx)-int(1)/(logx)dx+int(1)/((logx)^(2))dx ` Again using integration by parts, we get `xlog(logx)-(x)/(logx)-int(1)/((logx)^(2))dx+int(1)/((logx)^(2)dx+C` `=x[log(logx)-(1)/(logx)]+C` `therefore f(x)=log(logx),g(x)=(1)/(logx)` |
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| 321. |
Evaluate:`int{s in(logx)+cos(logx)}dx` |
| Answer» Correct Answer - ` x sin (log x) +c` | |
| 322. |
`int(2x-1)/(2x^2+2x+1)dx` |
| Answer» Correct Answer - `(1)/(2) log |2x^(2) +2x+1|-2 tan^(-1) (2x+1)+c` | |
| 323. |
Evaluate: `int(x^2+1)/(x^2-1) dx` |
| Answer» Correct Answer - `x+log |(x-1)/(x+1)|+c` | |
| 324. |
The value of `int(cos xdx)/((sinx-1)(sinx-2))` is equal toA. `log|(sinx-2)/(sinx-1)|+C`B. `log((sinx-1)/(sinx-2))+C`C. `log(sinx-2)+C`D. None of these |
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Answer» Correct Answer - A Let `l=int(cosxdx)/((sinx-1)(sinx-2))` Put `sinx = t rArr cos x dx=dt` `therefore" "l=int(dt)/((t-1)(t-2))=int((1)/(t-2)-(1)/(t-1))dt` `=log|t-2|-log|t-1|+C` `=log|(t-2)/(t-1)|+C=log|(sinx-2)/(sinx-1)|+C` |
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| 325. |
`(i) int(cos^(3) x+ sin^(3) x)/(sin^(2) x.cos ^(2) x)dx " "(ii) int(cos2x)/(cos^(2) x sin^(2)x) dx` |
| Answer» Correct Answer - `(i) sec x-cosec x+c " "(ii) -cot x-tan x+c` | |
| 326. |
`int(tanx)/(sqrt(sin^(4)x+cos^(4)x))dx` is equal toA. `ln(tan^(2)x)+sqrt(1+tan^(2)x)+C`B. `secx+C`C. `sqrt(1-tan^(2)x)+C`D. None of the above |
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Answer» Correct Answer - D `I=int(tanx)/(sqrt(sin^(4)x+cos^(4)x))dx=int(tanx.sec^(2)x)/(sqrt(1+tan^(4)x))dx` `"Put "tanx=t` `rArr" "sec^(2)x dx=dt` `int(tanx)/(sqrt(sin^(4)x+cos^(4)x))dx=int(t)/(sqrt(1+t^(4)))dt` `"Let "t^(2)=t rArr 2t dt = dy` `therefore" "I=int(dy)/(2sqrt(1+y^(2)))=(1)/(2)ln|y+sqrt(y^(2)+1)|+C` `=(1)/(2)ln|tan^(2)x+sqrt(tan^(4)x+1)|+C` |
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| 327. |
`int(3+2cosx)/((2+3cosx)^(2))dx` is equal toA. `((sinx)/(2+3cosx))+C`B. `((2cosx)/(2+3sinx))+C`C. `((2cosx)/(2+3cosx))+C`D. `((2sinx)/(2+3sinx))+C` |
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Answer» Correct Answer - A Divide numerator and denominator by `sin^(2)x`, then `int(("3 cosec"^(2)x+"2 cosec x cot x"))/(("2 cosec x "+3 cot x)^(2))dx` Put `" 2 cosec x "+"3 cot x = t` ` therefore" "(-"2 cosec x cot x "-"3 cosec"^(2)x)dx=dt` `=int-(dt)/(t^(2))=(1)/(t)+C=(1)/("2 cosec x"+"3 cot x")+C` `=(sinx)/((2+3 cos x))+C` |
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| 328. |
Find `int[log(logx)+1/((logx)^2)]dx` |
| Answer» Correct Answer - `x log (log x) -(x)/(log x) +c` | |
| 329. |
Evaluete `intx/(x^4+x^2+1)dx` |
| Answer» Correct Answer - `(1)/(sqrt(3))tan^(-1) ((2x^(2)-1)/(sqrt(3)))+c` | |
| 330. |
`" If " f(x) =int_(0)^(x)" t sin t dt tehn " f(x) is `A. x sin xB. x cos xC. sin x+x cos xD. |
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Answer» Correct Answer - b ` f(x) = int_(0)^(x) t sin t dt` Using partial integration `f(x) =[int sin t dt -int [((d)/(dt)t) int sin t dt ]dt]_(0)^(x)` `= [t( -cos t)]_(0)^(x) - int_(0)^(x) (-cos t)dt` `=[-t cos t + sin t]_(0)^(x)` `rArr f(x) =-x cos x + sin x` Now differebntiate with respect to x . f(x) = -[{x(-sin x)}+cos x] +cos x `= sin x- cos x + cos x =x sin x` |
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| 331. |
`intx/((x^2-a^2)(x^2-b^2))dx` |
| Answer» Correct Answer - `(1)/(2(a^(2)-b^(2)))log |(x^(2)-a^(2))/(x^(2)-b^(2))|+c` | |
| 332. |
If `I=int(sinx+sin^3x)/(cos2x) dx`=`Pcosx+Q log|f(x)|+R`,thenA. `P=(1)/(2), Q=(1)/(4sqrt2)`B. `P=(1)/(4),Q=-(1)/(sqrt2)`C. `f(x)=(cosx+1)/(sqrt2cosx-1)`D. `f(x)=(sqrt2cosx-1)/(sqrt2cosx+1)` |
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Answer» Correct Answer - D `l=int(sinx+sin^(3)x)/(cos2x)dx=int(sin x(2-cos^(2)x))/((2cos^(2)x-1))dx` Put `cos x= t rArr dx = -sin x dx` `therefore" "l=int(t^(2)-2)/(2t^(2)-1)dt=(1)/(2)int(2t^(2)-4)/(2t^(2)-1)dt` `=(1)/(2)int dt -(3)/(2)int (dt)/(2t^(2)-1)` `=(1)/(2)t-(3)/(2sqrt2)xx(1)/(2)ln|(sqrt2t-1)/(sqrt2t+2)|+C` `=(1)/(2)cos x-(3)/(4sqrt2)ln|(sqrt 2 cos x-1)/(sqrt2 cos x+1)|+C` So, `P=1//2, Q=(-3)/(4sqrt2)and f(x)=(sqrt2 cos x-1)/(sqrt2 cos x+1)` or `P=1//2, Q=(3)/(4sqrt2)and f(x)=(sqrt2 cos x+1)/(sqrt2 cos x-1)` |
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| 333. |
`int(x^(2)+cos^(2)x)/(x^(2)+1)."cosec"^(2)xdx` is equal toA. `cotx +cot^(-1)x+C`B. `-e^(ln tan^(-1))x-cot x+C`C. `C-cotx+cot^(-1)x`D. `-tan^(-1)x-("cosec x")/(sec x)+C` |
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Answer» Correct Answer - C `l=int(x^(2)+cos^(2)x)/(x^(2)+1)."cosec"^(2)xdx` `=int(x^(2)+1+cos^(2)x-1)/(x^(2)+1)."cosec"^(2)xdx` `=int(1-(sin^(2)x)/(x^(2)+1))."cosec"^(2)xdx` `=int("cosec"^(2)x-(1)/(x^(2)+1))dx` `=-cotx - tan^(-1)x+C_(1)=-cot x+cot^(-1)x-(pi)/(2)+C_(1)` `=-cotx+cot^(-1)x+C" where, "C=C_(1)-(pi)/(2)` |
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| 334. |
`int(dx)/(e^(x)+e^(-x)+2)` is equal toA. `(1)/(e^(x)+1)+C`B. `(1)/(e^(x)+1)+C`C. `(1)/(1+e^(-x))+C`D. None of these |
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Answer» Correct Answer - D `l=int(dx)/(e^(x)+e^(-x)+2)=int(e^(x)dx)/(e^(2x)+2e^(x)+1)=int(e^(x))/((e^(x)+1)^(2))dx` Put, `e^(x)+1=t` `therefore" "e^(x)dx=dt` `therefore" "l=int(dt)/(t^(2))=-(1)/(t)+C=(-1)/(e^(x)+1)+C` |
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| 335. |
The integral `int(dx)/(x^(2)(x^(4)+1)^(3//4))` equalA. `((x^(4)+1)/(x^(4)))^(1//4)+C`B. `(x^(4)+1)^(1//4)+C`C. `-(x^(4)+1)^(1//4)+C`D. `-((x^(4)+1)/(x^(4)))^(1//4)+C` |
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Answer» Correct Answer - D Let `l=int(dx)/(x^(2)(x^(4)+1)^((3)/(4)))=int(dx)/(x^(5)(1+(1)/(x^(4)))^((3)/(4)))` Put `1+(1)/(x^(4))=t^(4)" "rArr" "-(4)/(x^(5))dx=4t^(3)dt` `rArr" "(dx)/(x^(5))=-t^(3)dt` `therefore" "l=int(-t^(3)dt)/(t^(3))=-intdt=-t+C=-(1+(1)/(x^(4)))^((1)/(4))+c` |
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| 336. |
`intx/((1+sinx))dx` |
| Answer» Correct Answer -`x (tan x-sec x) +log (1+sin x) +c` | |
| 337. |
`int((x^(4)-x)^(1//4))/(x^(5))dx` is equal toA. `(4)/(15)(1-(1)/(x^(3)))^(5//4)+C`B. `(4)/(5)(1-(1)/(x^(3)))^(5//4)+C`C. `(4)/(15)(1+(1)/(x^(3)))^(5//4)+C`D. None of these |
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Answer» Correct Answer - A Let `l=int((x^(4)-x)^(1//4))/(x^(5))dx=int(x(1-(1)/(x^(3)))^(1//4))/(x^(5))dx` `=int((1-(1)/(x^(3)))^(1//4))/(x^(4))dx` Put `1-(1)/(x^(3))=t^(4) rArr (3)/(x^(4))dx= 4t^(3)dt` `therefore" "l=(4)/(3)int t.t^(3)dt=(4)/(3).((t^(5))/(5))+C=(4)/(!5)(1-(1)/(x^(3)))^(5//4)+C` |
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| 338. |
`int(2)/((2-x)^(2))root3((2-x)/(2+x))dx` is equal toA. `(4)/(3)((2+x)/(2-x))^(2//3)+C`B. `(3)/(4)((2+x)/(2-x))^(2//3)+C`C. `(3)/(4)((2-x)/(2+x))^(2//3)`D. `(3)/(4)((2+x)/(2-x))^(4//3)+C` |
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Answer» Correct Answer - B Put `(2-x)/(2+x)=z^(3) rArr x=(2-2z^(3))/(1+z^(3))` `rArr" "dx=(-12z^(2)dz)/((1+z^(3))^(2))` `therefore" "int(2)/((2-x)^(2)) root3((2-x)/(2+x))dx` `=int(2)/([2-(2-2z^(3))/(1+z^(3))]^(2)).z.((-12z^(3)))/((1+z^(3))^(2))dz` `=(-3)/(2)int(dz)/(z^(3))=(3)/(4z^(2))+C=(3)/(4)((2+x)/(2-x))^(2//3)+C` |
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| 339. |
The value of the integral `int(dx)/(x^(n)(1+x^(n))^(1//n)), n in N` isA. `(1)/((1-n))(1+(1)/(x^(n)))^(1-1//n)+C`B. `(1)/((1-n))(1-(1)/(x^(n)))^(1-1//n)+C`C. `(1)/((1-n))(1-(1)/(x^(n)))^(1-1//n)+C`D. `(1)/((1-n))(1-(1)/(x^(n)))^(1-1//n)+C` |
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Answer» Correct Answer - A `int(dx)/(x^(n)(1+x^(n))^(1//n))=int(dx)/(x^(n).x((1)/(x^(n))+1)^(1//n))` `=int(dx)/(x^(n+1)((1)/(x^(n))+1)^(1//n))` `"Put "(1)/(x^(n))+1=t` `rArr-(n)/(x^(n+1))dx=dt` `rArr" "-(1)/(n)int(dt)/(t^(1//n))=-(1)/(n)int t^(-1//n)dt=-(1)/(n).(t^(1-(1)/(n)))/((1-(1)/(n)))+C` `=(1)/((1-n))(1+(1)/(x^(n)))^(1-(1)/(n))+C` |
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| 340. |
The integral `int(dx)/((sqrtx+root3(x^(2))))` represents the functionA. `6{root3(x^(2))-root3(x)+ln|1+root3(x)|}+C`B. `3root3(x^(2))-6root6(x)+6ln|1+root6(x)|+C`C. `3root3(x^(2))+6root6(x)+6ln|1+root6(x)|+C`D. `6root3(x^(2))-3root3(x)+6ln|1+root3(x)|+C` |
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Answer» Correct Answer - B Let `l=int(dx)/((sqrtx+root3(x^(2))))` `"Put "x=t^(6)` `therefore" "dx=6t^(5)dt` `"Then, "l=int(6t^(5)dt)/((t^(3)+t^(4)))=6int(t^(2)dr)/((1+t))` `=6int(t-1+(1)/(1+t))dt` `=6{(t^(2))/(2)-t+ln|1+t|}+C` `=3t^(2)-6t+6ln|1+t|+C` `=3t^(2)-6t+6ln|1+t|+C` `=3.root3(x)-6.root6(x)+6ln|1+root6(x)|+C` |
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| 341. |
`int e^(sqrtx) dx` |
| Answer» Correct Answer - `2e^(sqrt(x))(sqrt(x)-1)+c` | |
| 342. |
If `d/(dx)f(x)=4x^3-3/(x^4)`such that `f(2)=0.`Then f(x) is(A) `x^4+1/(x^3)-(129)/8` (B) `x^3+1/(x^4)+(129)/8`(C) `x^4+1/(x^3)+(129)/8` (D) `x^3+1/(x^4)-(129)/8`A. `x^(4)+(1)/(x^(3))-(129)/(8)`B. `x^(3)+(1)/(x^(4))+(129)/(8)`C. `x^(4)+(1)/(x^(3))+(129)/(8)`D. `x^(3)+(1)/(x^(4))-(129)/(8)` |
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Answer» Correct Answer - A Given, `(d)/(dx)f(x)=4x^(3)-(3)/(x^(4))` `rArr" Anti - derivative of "(4x^(3)-(3)/(x^(4)))=f(x)` `therefore" "f(x)=int(4x^(3)-(3)/(x^(4)))dx=4intx^(3)dx-3 int x^(-4)dx` `rArr" "f(x)=4((x^(4))/(4))-3 ((x^(-3))/(-3))+C=x^(4)+(1)/(x^(3))+C` Also, `f(2)=0` `therefore" "f(2)=(2)^(4)+(1)/((2)^(3))+C` `rArr" "16+(1)/(8)+C=0 rArr C=-(16+(1)/(8))` `rArr" "C=-(129)/(8)` `therefore" "f(x)=x^(4)+(1)/(x^(4))-(129)/(8).` |
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| 343. |
`int (dx)/(x(x^(2)+1))" equals "`A. `log|x|+(1)/(2)log(x^(2)+1)+C`B. `-log|x|+(1)/(2)log(x^(2)+1)+C`C. `(1)/(2)log|x|+log (x^(2)+1)+c`D. |
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Answer» Correct Answer - A `" Let " (1)/(x(x^(2) +1))+(A)/(x)+(Bx+c)/(x^(2)+1)` 1=A `(x^(2) +1) +(Bx+c)x` x=0 then 1=A (0+1) +0 `rArr ` A=1 Equating the coefficient of `x^(2)` `0= A+B rArr B=- A=-1` Equating the coefficeints of `x,0 =C` ` :. (1)/(x(x^(2) +1)) =(1)/(x)+ (-x)/(x^(2) +1)` ` :. int(1)/(x(x^(2)+1))dx = int ((1)/(x)-(x)/(x^(2)-1))dx` `=log |x|-(1)/(2) log (x^(2)+1)+c` |
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| 344. |
Integral of `(x^(3)+3x+4)/(sqrtx)` isA. `(2)/(7)x^(5//2)+(2)/(3)x^(3//2)+8x^(1//2)+C`B. `(2)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C`C. `(1)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C`D. `x^(7//2)+3x^(3//2)+4x^(1//2)+C` |
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Answer» Correct Answer - B `int(x^(3)+3x+4)/(sqrtx)dx`. `int(x^(3)+3x+4)/(sqrtx)dx=int(x^(3))/(sqrtx)dx+3int(x)/(sqrtx)dx+4int(1)/(sqrtx)dx` `=int x^(3).x^(-1//2)dx+3 intx^(1//2)dx+4 int x^(-1//2)dx` `=int x^(5//2)dx+3int x^(1//2)dx+4 intx^(-1//2)dx` `=(x^((5//2)+1))/((5//2)+1)+(3x^((1//2)+1))/((1//2)+1)+(4x^((-1//2)+1))/((-1//2)+1)+C` `" "[because int x^(n)dx=(x^(n+1))/(n+1)]` `=(x^(7//2))/(7//2)+(3x^(3//2))/(3//2)+(4x^(1//2))/(1//2)+C` `=(2)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C` |
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| 345. |
`"The integral " int(dx)/(x^(2)(x^(4)+1)^(3//4))" equals"`A. `(1+x^(4)) ^(1//4)+c`B. `(1-x^(-4))^(1//4) +c`C. `-(1+x^(-4))^(1//4)+c`D. |
| Answer» Correct Answer - D | |
| 346. |
Evaluate :`int_(-pi//4)^(pi//4) |sin x|dx` |
| Answer» Correct Answer - `(2-sqrt(2))` | |
| 347. |
Evaluate :`int_(0)^(8) |x-5|dx` |
| Answer» Correct Answer - 17 | |
| 348. |
Evaluate :` int_(-pi//2)^(pi//2) |sin x|dx` |
| Answer» Correct Answer - 2 | |
| 349. |
Evaluate : `int_(0)^(1) (1-x)^(3//2) dx` |
| Answer» Correct Answer - `(4)/(35)` | |
| 350. |
Evaluate : `int_(0)^(4) x(4-x)^(3//2)dx` |
| Answer» Correct Answer - `(512)/(35)` | |