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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.
| 151. |
Evaluate:`int(dx)/((a^2+x^2)^(3/2))` |
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Answer» Put `x=a tan theta implies dx = a sec^(2) theta d theta` ` :. I= int (dx)/(a^(2)+x^(2))^(3//2)` `= int(a sec^(2)theta)/((a^(2)+a^(2) tan^(2)theta)^(3//2))d theta` `=int(a sec^(2)theta)/(a^(3)(sec^(2)theta)^(3//2))d theta` `=(1)/(a^(2))int(d theta)/(sec theta)` `=(1)/(a^(2))int cos theta d theta` `=(1)/(a^(2))sin theta +c` Now `tan theta=(x)/(a)` ` :. sin theta = (x)/(sqrt(x^(2)+a^(2)))` ` :. I= (x)/(a^(2)sqrt(x^(2)+a^(2)))+c` |
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| 152. |
Evaluate:`int1/((x^2+2x+2)^2)dx` |
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Answer» `I=(1)/((x^(2)+2x+2)^(2))dx` `=(1)/([(x+1)^(2)+1^(2)]^(2))dx` Put `x+1=tan theta` or `dx=sec^(2)theta d theta` ` :. I= int(1)/((tan^(2)theta+1)^(2))sec^(2)theta d theta` `=int cos^(2)theta d theta` `=(1)/(2)int (1+cos2theta)d theta` `=(1)/(2)(theta+(sin 2theta)/(2))+C ` `=(1)/(2)(theta +sin theta cos theta)+C` `=(1)/(2){tan^(-1)(x+1)+(x+1)/(x^(2)+2x+2)}+C` |
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| 153. |
For `x >1,int sin^(- 1)((2x)/(1+x^2))dx` is equal toA. `x tan^(-1)x-In|sec(tan^(-1)x)|+c`B. `x tan^(-1)x+In|sec(tan^(-1)x)|+c`C. `x tan^(-1)x-In|cos(tan^(-1)x)|+c`D. none of these |
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Answer» Correct Answer - D `I= int sin^(-1)((2x)/(1+x^(2)))dx` `"Let " x=tan theta` `"or " dx=sec^(2)theta d theta ` ` :. I=intsin^(-1)((2tan theta)/(1+tan^(2)theta))sec^(2) theta d theta ` ` =2 int theta sec^(2)thetad theta ` `=2(theta tan theta-In|sec theta|)+C` `=2(xtan^(-1)x-In|sec(tan^(-1)x)|)+C` |
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| 154. |
Let `I_n=int tan^n x dx, (n>1)`. If`I_4+I_6=a tan^5 x + bx^5 + C`, Where `C` is a constant of integration, then the ordered pair `(a,b)` is equal to :A. `(-(1)/(5),1)`B. `((1)/(5),0)`C. `((1)/(5),-1)`D. `(-(1)/(5),0)` |
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Answer» Correct Answer - B We have, `I_(n)=int tan^(n)x dx` `therefore I_(n)+I_(n+2)=int tan^(n)xdx + int tan^(n+2)xdx` `" "=int tan^(n)x(1+tan^(2)x)dx` `" "=int tan^(n) x sec^(2)xdx=(tan^(n+1))/(n+1)+C` Put n = 4, we get `I_(4)+I_(6)=(tan^(5)x)/(5)+C` `therefore" "a=(1)/(5)and b = 0` |
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| 155. |
Evaluate:`int(1/(x3+x4)+(ln(1+x6)/(x3+sqrt(x)))dx` |
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Answer» Correct Answer - `(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+(4)/(3)x^(1//2)-(12)/(5)x^(5//12)+(1)/(2)x^(1//3)-4x^(1//4)-7x^(1//6)-12x^(1//12)+(2x^(1//2)-3x^(1//3)+6x^(1//6)+11)"In"(1+x^(1//6))+12"In"(1+x^(1//2))-3["In(1+x^(1//6))]^(2)+c` Let `I=int ((1)/(root(3)(x)+root(4)(x))+(In(1+root(6)(x)))/(root(3)(x)+sqrt(x)))dx` `therefore" "I=I_(1)+I_(2)` where, `I_(1)=int((1)/(root(3)(x)+root(4)(x)))dx`, `I_(2)=int(In(1+root(6)(x)))/(root(3)(x)+root(4)(x))dx` Now, `I_(1)=int((1)/(root(3)(x)+root(4)(x)))dx` Put `x=t^(12) rArr dx = 12 t^(11)dt` `therefore" "I_(1)=12 int(t^(11))/(t^(4)+t^(3t))d` `" "=12 int(t^(8)dt)/(t+1)` `" "=12 int(t^(7)-t^(6)+t^(5)-t^(4)+t^(3)-t^(2)+t-1)dt + 12 int(dt)/(t+1)` `" "=12((t^(8))/(8)-(t^(7))/(7)+(t^(6))/(6)-(t^(5))/(5)+(t^(4))/(4)-(t^(3))/(3)+(t^(2))/(2)-t)+12 In (t+1)` and `I_(2)=int{(In(1+root(6)(x)))/(root(3)x+sqrt(x))}dx` Put `x = u^(6) rArr dx = 6 u^(5) du` `therefore I_(2)=int(In(1+u))/(u^(2)+u^(3))6u^(5)du=int(In(1+u))/(u^(2)(1+u))*6u^(5)du` `=6int(u^(3))/((u+1))In(1+u)du` `=6int((u^(3)-1+1)/(u+1))In(1+u)du` `=6int(u^(2)-u+1-(1)/(u+1))In(1+u)du` `=6 int(u^(2)-u+1)In(1+u)du-6int(In(1+u))/((u+1))du` `=6((u^(3))/(3)-(u^(2))/(2)+u)In(1+u)-int(2u^(3)-3u^(2)+6u)/(u+1)du-6(1)/(2)[In(1+u)]^(2)` `=(2u^(3)-3u^(2)+6u)In(1+u)-int(2u^(2)-5u+(11u)/(u+1))du-3[In(1+u)]^(2)` `=(2u^(3)-3u^(2)+6u)In(1+u)-((2u^(3))/(3)-(5)/(2)u^(2)+11u-11 In(u+1))-3[In(1+u)]^(2)` `therefore" "I=(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+2x^(1//2)-(12)/(5)x^(5//12)+3x^(1//3)-4x^(1//4)` `-6x^(1//6)-12x^(1//12)+12 In(x^(1//12)+1)+(2x^(1//2)-3x^(1//3)+6x^(1//6))In(1+x^(1//6))` `-[(2)/(3)x^(1//2)-(5)/(2)x^(1//3)11x^(1//6)-11 In(1+x^(1//6))]-3[In(1+x^(1//6))]+c` `=(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+(4)/(3)x^(1//2)-(12)/(5)x^(5//12)+(1)/(2)x^(1//3)-4x^(1//4)-7x^(1//6)-12x^(1//12)` `+(2x^(1//2)-3x^(1//3)+6x^(1//6)+11)In(1+x^(1//6))+12 In (1+x^(1//12))-3[In(1+x^(1//6))]^(2)+c` |
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| 156. |
Evaluate:`int1/xsqrt((1-sqrt(x))/(1+sqrt(x))dx)` |
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Answer» Put ` x= cos^(2)theta` `impliesdx= -2cos theta sin theta d theta` ` :. I=int(1)/(x)sqrt((1-sqrt(x))/(1+sqrt(x)))dx` `=intsqrt((1-cos theta)/(1+cos theta))* (-2cos theta sin theta)/(cos^(2) theta) d theta` `= -int ("sin"(theta)/(2))/("cos"(theta)/(2))*(4"sin"(theta)/(2)"cos"(theta)/(2))/("cos"theta)d theta ` `= -2 int (2 "sin"^(2)(theta)/(2))/(cos theta) d theta` `= -2int(1-cos theta)/(cos theta)d theta` `= -2 int (sec theta -1)d theta` `= - 2 [log_(e)|sec theta+tan theta|-theta] +c` `= -2[log_(e)|(1)/(sqrt(x))+(sqrt(1-x))/(sqrt(x))|-"cos"^(-1)sqrt(x)]+c` `= -2[log_(e)|(1+sqrt(1-x))/(sqrt(x))|-cos^(-1)sqrt(x)] +c` |
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| 157. |
Evaluate : `intsqrt(xsqrt(x^(2)+2))`dx |
| Answer» `1/3x+sqrt(x^(2)+2)^(3//2)-2/(x+sqrt(x^(2)+2)^(1//2)+C` | |
| 158. |
Evaluate `int(dx)/((x^3+3x^2+3x+1)sqrt(x^2+2x-3))` |
| Answer» `sqrt(x^(2)+2x-3)/(8(x+1)^(2))+1/16.cos^(-1)(2/(x+1))+C` | |
| 159. |
evaluate `int(cos2x-3)/(cos^4sqrt(4-cot^2x)).dx` |
| Answer» `C-1/3tanx.(2+tan^(2)x).sqrt(4-cot^(2)x)` | |
| 160. |
If `f(x)=inte^(x)(tan^(-1)x+(2x)/((1+x^(2))^(2)))dx,f(0)=0` then the value of f(1) isA. `e((pi)/(4)-(1)/(2))+1`B. `e((pi)/(4)+(1)/(2))+1`C. `e((pi)/(2)-(1)/(4))+1`D. `e^(-1)((pi)/(4)-(1)/(2))+1` |
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Answer» Correct Answer - A If `inte^(x)(tan^(-1)x+(2x)/((1+x^(2))^(2)))dx` `=inte^(x)(tan^(-1)x-(1)/(1+x^(2))+(1)/(1+x^(2))+(2x)/((1+x^(2))^(2)))dx` `=e^(x)(tan^(-1)x-(1)/(1+x^(2)))+c` `f(0)=0rArr c=1` `rArr" "f(1)=e((pi)/(4)-(1)/(2))+1` |
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| 161. |
The value of `int(dx)/(cos^3 sqrtsin2x)` is equal toA. `sqrt(2)(cossqrt(cosx)+1/2tan^(5//2)x)+C`B. `sqrt(2)(sqrt(tanx)+1/5tan^(5//2)x)+C`C. `sqrt(2)(sqrt(tanx)-1/5tan^(5//2)x)+C`D. `sqrt(2)(sqrt(cosx)-1/5tan^(5//2)x)+C` |
| Answer» Correct Answer - B | |
| 162. |
Evaluate: `int(dx)/(a+bcosx)^(2), (a gt b)` |
| Answer» `-(bsinx)/((a^(2)-b^(2))(a+bcosx))+(2a)/((a^(2)-b^(2))^(3//2)tan^(-1)sqrt((a-b)/(a+b))tanx/2+C` | |
| 163. |
Evaluate: `intsqrt(2-x-x^(2))/(x^(2))`dx |
| Answer» `-sqrt(2-x-x^(2))/(x)+sqrt(2)/4"ln"|(4-x+2sqrt(2)sqrt(2-x-x^(2))/(x)-sin^(-1))|(2x+1)/(3)+K` | |
| 164. |
Evaluate: `int(3-4x)/(2x^(2)-3x+1)dx` |
| Answer» `-"ln"|(2x^(2)-3x+1)|+C` | |
| 165. |
`int((cosx)^(n-1))/((sinx)^(n+1))dx=` (A) `-cot^n x/n+c` (B) `-cot^n x/(n+1)+c` (C) `cot^n x/n+c` (D) `cot^n x/(n+1)+c`A. `(cot^(n)x)/(n)`B. `(-cot^(n-1)x)/(n-1)`C. `(-cot^(n)x)/(n)`D. `(cot^(n-1)x)/(n-1)` |
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Answer» Correct Answer - C `int(cos^(n-1)x)/(sin^(n+1)x)dx=int cot^(n-1)x" cosec"^(2)xdx` `=(-cot^(n)x)/(n)+C` |
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| 166. |
Evaluate: `int(x-1)sqrt(1+x+x^(2))`dx |
| Answer» `1/3(x^(2)+x+1)^(3//2)-3/8(2x+1)sqrt(1+x+x^(2))-9/16log(2x+1+2sqrt(x^(3)+x+1))+C` | |
| 167. |
Evaluate: `int(sin^8x-cos^8x)/(1-2sin^2x+cos^2x) dx` |
| Answer» Correct Answer - `-1/2sin2x+C` | |
| 168. |
The integral `intcos(log_(e)x)dx` is equal to: (where C is a constant of integration)A. `x[cos(log_(e)x)-sin(log_(e)x)]+C`B. `x/2[sin(log_(e)x)-cos(log_(e)x)]+C`C. `x[cos(log_(e)x)+sin(log_(e)x)]+C`D. `x/2[cos(log_(e)x)+sin(log_(e)x)]+C` |
| Answer» Correct Answer - 4 | |
| 169. |
If `int x^(5)e^(-x^(2))dx = g(x)e^(-x^(2))+C`, where C is a constant of integration, then g(-1) is equal toA. -1B. 1C. `-(1)/(2)`D. `-(5)/(2)` |
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Answer» Correct Answer - D Let given integral, `I = int x^(5) e^(-x^(2)) dx` Put `x^(2) = t rArr 2xdx = dt` So, `I=(1)/(2)int t^(2)e^(-1)dt` `=(1)/(2)[(-t^(2)e^(-t))+inte^(-t)(2t)dt]" "["Integration by parts"]` `=(1)/(2)[-t^(2)e^(-t)+2t(-e^(-t))+int 2e^(-t)dt]` `=(1)/(2)[-t^(2)e^(-t)-2te^(-t)-2e^(-t)]+C` `=-(e^(-t))/(2)(t^(2)+2t+2)+C` `=-(e^(-x^(2)))/(2)(x^(4)+2x^(2)+2)+C" "[therefore t = x^(2)]" "....(i)` `therefore` It is given that, `I=int x^(5)e^(-x^(2))dx = g(x)*e^(-x^(2))+C` By Eq. (i), comparing both sides, we get `g(x)=-(1)/(2)(x^(4)+2x^(2)+2)` So, `g(-1)=-(1)/(2)(1+2+2)=-(5)/(2)` |
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| 170. |
`int((cosx)^(n-1))/((sinx)^(n+1))dx=` (A) `-cot^n x/n+c` (B) `-cot^n x/(n+1)+c` (C) `cot^n x/n+c` (D) `cot^n x/(n+1)+c`A. `(cot^(n)x)/(n)`B. `(-cot^(n-1)x)/(n-1)`C. `(-cot^(n)x)/(n)`D. `(cot^(n-1)x)/(n-1)` |
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Answer» Correct Answer - C `int(cos^(n-1)x)/(sin^(n+1)x)dx=int cot^(n-1)x" cosec"^(2)xdx` `=(-cot^(n)x)/(n)+C` |
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| 171. |
What is `int (dx)/(sin^(2) x cos^(2) x)` equal to ? where c is the constant of integrationA. `tan x + cot x + c`B. `tan x - cot x + c`C. `(tan x + cot x)^(2) + c`D. `(tan x - cot x)^(2) + c` |
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Answer» Correct Answer - B Let `I = int (dx)/(sin^(2) x cos^(2) x) = int ((sin^(2) x + cos^(2) x)dx)/(sin^(2) x cos^(2) x)` `= int [(sin^(2)x)/(sin^(2) x cos^(2) x) + (cos^(2)x)/(sin^(2) x cos^(2) x)] dx` `= int [(1)/(cos^(2)x) + (1)/(sin^(2)x)] dx` `= int (sec^(2) x + "cosec"^(2)x) dx` `= tan x - cot x + c` where c is constant of Integration |
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| 172. |
If `int e^(3x)cos4xdx=e^(3x)(Asin4x+Bcos4x)+C`, then:A. 4A=3BB. 2A=3BC. 3A=4BD. 4A+3B=1 |
| Answer» Correct Answer - CD | |
| 173. |
`int 5^(x+tan^(-1)x)*((x^(2)+2)/(x^(2)+1))dx`. |
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Answer» Let `x+tan^(-1)x=t` ` :. (1+(1)/(x^(2)+1))dx=dt " or "((x^(2)+2)/(x^(2)+1))dx=dt` ` :. int 5^(x+tan^(-1)x)*((x^(2)+2)/(x^(2)+1))dx=int 5^(t) dt` `=(5^(t))/(log_(e)5)+c` `=(5^(x+tan^(-1)x))/(log_(e)5)+c` |
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| 174. |
Consider the following : 1. `int l n 10 dx = x + c` 2. `int 10^(x) dx = 10^(x) + c` where c is the constant of integration. Which of the above is/are correct ?A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2 |
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Answer» Correct Answer - D (1) Let `I = int l n 10dx = l n 10 int dx = [l n 10] x + c` (2) Let `I = int 10^(x) dx = (10^(x))/(log_(e) 10) + c = (10^(x))/(l n 10) + c` |
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| 175. |
Evaluate: `inttan^(-1)x."ln"(1+x^(2))dx` |
| Answer» `xtan^(-1)x."ln"(1+x^(2))+(tan^(-1)x)^(2)-2xtan^(-1)x+"ln"|sinx+cosx|+C` | |
| 176. |
`"o f"int(dx)/(x^2(x^4+1)^(3/4))`A. `((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 `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` Hence, the integral becomes `int(-t^(3)dt)/(t^(3))=-int dt =-t+c=-(1+(1)/(x^(4)))^(1//4)+c` |
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| 177. |
If `f(x)=(x)/((1+x^(n))^(1//n)) "for n" ge 2 and g(x) = underset("f occurs n times")ubrace(("fofo....of"))(x). "Then",intx^(n-2)g(x) dx` equalA. `(1)/(n(n-1))(1+nx^(n))^((1-(1)/(n))+c`B. `(1)/(n-1)(1+nx^(n))^((1-(1)/(n))+c`C. `(1)/(n(n+1))(1+nx^(n))^((1+(1)/(n))+c`D. `(1)/(n+1)(1+nx^(n))^((1+(1)/(n))+c` |
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Answer» Correct Answer - A Given, `f(x)=(x)/((1+x^(n))^(1//n))"for n" ge 2` `therefore" "ff(x)=(f(x))/([1+f(x)^(n)]^(1//n))=(x)/(1+2x^(n))^(1//n) and fff(x) = (x)/((1+3x^(n))^(1//n))` `therefore" "g(x)=underset("n times")ubrace("fofo....of")(x)=(x)/((1+nx^(n))^(1//n))` `"Let"" "I=int x^(n-1)g(x)dx = int(x^(n-1)dx)/((1+nx^(n))^(1//n))` `" "=(1)/(n^(2))int(n^(2)x^(n-1)dx)/((1+nx^(n))^(1//n))=(1)/(n^(2))int((d)/(dx)(1+nx^(n)))/((1+nx^(n))^(1//n))dx` `" "I=(1)/(n(n-1))(1+nx^(n))^(1-(1)/(n))+c` |
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| 178. |
Evaluate:`int(e^(sqrt(x))cos(e^(sqrt(x))))/(sqrt(x))dx` |
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Answer» Put `e^(sqrt(x))=t" or " (e^(sqrt(x)))/(sqrt(x))dx =2dt` ` :. int(e^(sqrt(x))cos(e^(sqrt(x))))/(sqrt(x))dx=2intcost dt` `=2 sin t +c` `=2sin e^(sqrt(x))+c` |
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| 179. |
Evaluate: `int 2^(2^(2^x)) * 2^(2^x) * 2^x dx` |
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Answer» `I=int2^(2^(2^(x)))2^(2^(x))2^(x) dx` Let `2^(2^(2^(x)))=t` or `2^(2^(2^(x)))2^(2^(x))2^(x)(log2)^(3)dx=dt` ` :. I=int(1)/((log2)^(3))dt=(1)/((log2)^(3))t+c` `=(1)/((log2)^(3))2^(2^(2^(x)))+C` |
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| 180. |
Find`int(e^x(1+x))/(cos^2(x e^2))dx` |
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Answer» Let `z=xe^(x). " Then " dz=(1e^(x)+xe^(x))dx=e^(x)(1+x)dx`. Thus, `int(e^(x)(1+x))/(cos^(2)(xe^(x)))dx=int(dx)/(cos^(2)z)` `=int sec^(2)zdz` `=tan z+c=tan(xe^(x))+c` |
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| 181. |
`int (" In" x)^(-1) dx - int ("In" x)^(-2) dx` is equal toA. `x ("In " x)^(-1) + c`B. `x("In " x)^(-2) + c`C. `x(" In "x) +c`D. `x("In " x)^(2) + c` |
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Answer» Correct Answer - A `int (l nx)^(-1) dx - int (l n x)^(-2) dx` `= int [(1)/(l nx) - (1)/((l nx)^(2))].dx` Put `l nx = t rArr x = e^(t)` `dx = e^(t).dt` `:. int [(1)/(l nx) - (1)/((l nx)^(2))] dx = int ((1)/(t) - (1)/(t^(2))) .e^(t).dt` `= int e^(t) .((1)/(t) - (1)/(t^(2))) .dt` `= (e^(t))/(t) + c = (x)/(l nx) + c` `= x(l nx)^(-1) + c` |
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| 182. |
What is `int (x^(2) + 1)^((5)/(2))x dx` equal to ? where c is a constant of integrationA. `(x^(2) + 1)^((7)/(2)) + C`B. `(2)/(7) (x^(2) + 1)^((7)/(2)) + c`C. `(1)/(7) (x^(2) + 1)^((7)/(2)) + c`D. None of the above |
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Answer» Correct Answer - C Let `I = int x (x^(2) + 1)^(5//2) dx` Put `x^(2) + 1 = t` `2xdx = dt` `xdx = (dt)/(2)` `:. I = int (t)^(5//2) (dt)/(2) = (1)/(2) ((t^(7//2))/(7//2)) + c = (1)/(7) (x^(2) + 1)^(7//2) + c` |
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| 183. |
What is `int a^(x) e^(x) dx` equal to ? where c is the constant of integrationA. `(a^(x)e^(x))/(l na) + c`B. `a^(x) e^(x) + c`C. `(a^(x) e^(x))/(l n(ae)) + c`D. None of the above |
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Answer» Correct Answer - C Let `I = int underset(I)a^(x) underset(II)(e)^(x) dx` `I = a^(x) int e^(x) dx - int a^(x) l n a e^(x) dx` `I = a^(x) e^(x) - l n a int a^(x) e^(x) dx` `I = a^(x) e^(x) - l n a. (I)` `rArr (1 + l n a) I = a^(x) e^(x)` `rArr I = (a^(x) e^(x))/(l n (ae")) + c ( :. l n e = 1)` |
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| 184. |
Evaluate `int(sin^(6)x)/(cos^(8)x)dx` |
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Answer» Correct Answer - `(tan^(7)x)/(7)+C` `int(sin^(6)x)/(cos^(8)x)dx=int(sin^(6)x)/(cos^(6)x)xx(1)/(cos^(2)x)dx` `=inttan^(6)x sec^(2)xdx` `=(tan^(7)x)/(7)+C` |
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| 185. |
Evaluate: `int1/(x^(1//2)+x^(1//3)) dx` |
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Answer» `(1)/(x^(1//2)+x^(1//3))=(1)/(x^(1//3)(1+x^(1//6)))` Let `x=t^(6)impliesdx=6t^(5) dt` ` :. int(1)/(x^(1//2)+x^(1//3))dx=int(1)/(x^(1//3)(1+x^(1//6)))dx` `=int(6t^(5))/(t^(2)(1+t))dt` `=6int(t^(3))/((1+t))dt` On dividing, we obtain `int(1)/(x^(1//2)+x^(1//3))dx=6int{(t^(2)-t+1)-(1)/(1+t)}dt` `=6[((t^(3))/(3))-((t^(2))/(3))+t-log|1+t|]+C` `=2x^(1//2)-3x^(1//3)+6x^(1//6)-6 log(1+x^(1//6))+C` |
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| 186. |
What is `int ((1)/(cos^(2)x) - (1)/(sin^(x)x))dx` equal to ? where c is the constant of integrationA. `2 "cosec" 2x + c`B. `-2 cot 2x + c`C. `2 sec 2x + c`D. `-2 tan 2x + c` |
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Answer» Correct Answer - A Let `I = int ((1)/(cos^(2)x) - (1)/(sin^(2)x)) dx` `= int (sec^(2) x - "cosec"^(2) x) dx` `= int sec^(2)x dx - int "cosec"^(2) x dx` `tan x + cot x + c` `= tan x + (1)/(tan x) + c = (tan^(2) x+ 1)/(tan x) + c = (sec^(2)x)/(tan x) + c` `= (2)/(2 sin x cos x) + c = (2)/(sin 2x) + c = 2 "cosec" 2x + c` |
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| 187. |
Evaluate: `int(8x-11)/sqrt(5+2x-x^(2))`dx |
| Answer» `-8 sqrt(5+2x-x^(2))-3sin^(-1)(x-1)/sqrt(6)+C` | |
| 188. |
Evaluate `int(dx)/((1+sinx)^(1//2))` |
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Answer» Correct Answer - `sqrt(2) log|(tan ((x)/(4)+(pi)/(x)))|+C` `int(dx)/((1+sinx)^(1//2))=int(dx)/("cos"(x)/(2)+"sin"(x)/(2))` `=(1)/(sqrt(2))int(dx)/("sin"((x)/(2)+(pi)/(4)))` `=(1)/(sqrt(2))int"cosec"((x)/(2)+(pi)/(4))dx` `=(1)/(sqrt(2))(log|tan((x)/(4)+(pi)/(8))|)/((1)/(2))+C` `=sqrt(2) log|(tan ((x)/(4)+(pi)/(x)))|+C` |
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| 189. |
What is `int e^(l nx) dx` equal to ? where c is constant of integrationA. `xe^(l nx) + c`B. `-xe^(-l nx) + c`C. `x + c`D. `(x^(2))/(2) + c` |
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Answer» Correct Answer - D Let `I = int e^(l nx) dx = int x dx = (x^(2))/(2) + c` |
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| 190. |
Evaluate:`int(x^2-1)/((x^4+3x^2+1)tan^(-1)(x+1/x))dx` |
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Answer» `I=int(x^(2)-1)/((x^(4)+3x^(2)+1)tan^(-1)(x+(1)/(x)))dx` `=int(1-(1)/(x^(2)))/((x^(2)+(1)/(x^(2))+3)tan^(-1)(x+(1)/(x)))dx` Put `x+(1)/(x)=t` ` :. (1-(1)/(x^(2)))dx=dt " and " x^(2)+(1)/(x^(2))+3=t^(2)+1` ` :. I=int(dt)/((t^(2)+1)tan^(-1)t)="In"|tan^(-1)t|+C` `="In"|tan^(-1)(x+(1)/(x))|+C` |
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| 191. |
Evaluate : `int(1/(2x^(2)+x-1))`dx |
| Answer» `1/3"ln"|(2x-1)/(2x+2)|+C` | |
| 192. |
Evaluate: `int(sinx)/(sin(x-a)) dx` |
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Answer» Correct Answer - `(x-a)cosa+sin a log|sin(x-a)|+C` `int(sinx)/(sin(x-a))dx=int(sin(x-a+a))/(sin(x-a)) dx` `=int(sin(x-a)cosa+cos(x-a)sina)/(sin(x-a))dx` `=cos a int dx+sin a int(cos(x-a))/(sin(x-a))dx` `=(cosa)x+sina|log sin(x-a)|+C` `=(x-a)cosa+sin a log|sin(x-a)|+C` |
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| 193. |
Evaluate `int(dx)/(cot^(2)x-1)` |
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Answer» Correct Answer - `(1)/(4)log_(e)|sec 2x+tan 2x|-(x)/(2)+C` `int(dx)/(cot^(2)x-1)=int(sin^(2)x dx)/(cos 2x)` `=(1)/(2)int((1-cos2x)dx)/(cos2x)` `=(1)/(2)[int sec 2x dx-int dx]` `=(1)/(2)[(1)/(2)log_(e)|sec 2x +tan 2x| -x]+C` `=(1)/(4)log_(e)|sec 2x+tan 2x|-(x)/(2)+C` |
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| 194. |
What is `int (l nx)/(x) dx` equal to ?A. `((l n x)^(2))/(2) + c`B. `((l nx))/(2) + c`C. `(l nx)^(2) + c`D. None of the above |
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Answer» Correct Answer - A Let `I = int (l nx)/(x) dx` Put `l n x = t` `(1)/(x) dx = dt` `:. I = int t dt = (t^(2))/(2) + c` where c is the constant of integration `= ((l nx)^(2))/(2) + c` |
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| 195. |
Ecaluate the following: (i) `int(sec^(2)x)/(3+tanx)dx " (ii) " int(e^(x)-e^(-x))/(e^(x)+e^(-x))dx` (iii) `int(1-tanx)/(1+tanx)dx " (iv) " int(1)/(1+e^(-x))dx` |
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Answer» (i) `int(sec^(2)x)/(3+tanx)dx=int((d)/(dx)(3+tanx))/(3+tanx)dx` `=log|3+tanx|+C` (ii) `int(e^(x)-e^(-x))/(e^(x)+e^(-x))dx=int((d)/(dx)(e^(x)+e^(-x)))/(e^(x)+e^(-x))dx` `=log_(e)(e^(x)+e^(-x))+C` (iii) `int(1-tanx)/(1+tanx)dx =int(cosx-sinx)/(cosx+sinx)dx` `=int((d)/(dx)(cosx+sinx))/(cosx+sinx)dx` `=log_(e)|cosx+sinx|+C` (iv) `int(1)/(1+e^(-x))dx=int(e^(x))/(e^(x)+1)dx` `=int((d)/(dx)(e^(x)+1))/(e^(x)+1)dx` `=log_(e)(1+e^(x))+C` |
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| 196. |
Evaluate: `intx^(2)e^(x)`dx |
| Answer» `x^(3)-2xe^(x)+2e^(x)+C` | |
| 197. |
Evaluate `int tan^(3)x dx` |
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Answer» Correct Answer - `(1)/(2) tan^(2)x-log|sec x|+C` `I=int tan^(3)x dx` `=int tan^(2)x tanx dx` `=int (sec^(2)x-1)tanx dx` `=int tanx sec^(2) x dx - int tanx dx` `=I_(1)-log|secx|+C, " where " I_(1)=int tanx sec^(2)x dx` Putting `tan x =t " and " sec^(2) x dx=dt " in " I_(1),` we get `I_(1)= int t dt = (t^(2))/(2)=(1)/(2) tan^(2)x +C` Hence, `I=(1)/(2) tan^(2)x-log|sec x|+C`. |
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| 198. |
Resolve `1/(2x^(3)+3x^(2)-3x-2)` into partial fractions: |
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Answer» We have, `1/(2x^(3)+3x^(2)-3x-2)=1/((x-1)(x+2)(2x+1))` Let `1/(2x^(2)+3x^(2)-3x-2)=A(x-1)+B/(x+2)+C/(2x1)`. Then, `rArr 1=A(x+2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x+)`…………..(i) Putting `x-1=0` or `x=1` in (i), we get `rArr A=1/9` Putting `x=-2`, in (i), we obtain `B=1/9` Putting `x=-1/2`in (i), we obtain `C=-4/9` `therefore 1/(2x^(3)+3x^(2)-3x-2)=1/((x-1)(x+2)(2x+1))=1/(9(x-1))+1/(9(x+2))-4/(9(2x+1))` |
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| 199. |
Evaluate : `int(3x+1)/((x-1)^(3)(x+1))dx` |
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Answer» Let `(3x+1)/((x-1)^(3)(x+1))= A(x+1)+B/(x-1)+C/(x-1)^(2)+D/(x-1)^(3)`…………(i) Multiplying both sides by (x+1) and then putting `x=-1`, we get `A=(-2)/(-2)^(3)=1/4` Multiplying both sides by `(x-1)^(3)` and then pulling x=1, wee get `D=4/2-2` From (i), we get `3x+1=A(x-1)^(3)+B(x-1)^(2)(x+1)+C(x-1)(x+1)+D(x+1)` `1=-A+B-C+D` `rArr 1=-1/4+B-C+2 rArr B-C=-3/4` Putting x=2, we get `7=A+3B+3C+3D` `rArr 7=1/4+3B+3C+6 rArr 3B+3C=3/4 rArr B+C=1/4` Solving B+C`=1/4` and `B-C=-3/4`, we get `B=-1/4, C=1/2` Substituting the values of A,B,C and D in (i), we get `rArr (3x+1)/((x-1)^(3)(x+1))=1/4.(1/(x+1))-1/(4(x-1))+1/(2(x-1)^(2))+2/(x-1)^(3)` `rArr int(3x+1)/((x-1)^(3)(x+1))dx=1/4int1/(x+1)dx-1/4int1/(x-1)dx+1/2int1/(x-1)^(2)dx+2int1/(x-1)^(3)`dx `=1/4ln|x+1|-1/4ln|x-1|-1/(2(x-1))-1/(2(x-1))-1/(x-1)^(2)+C` |
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| 200. |
What is `int (dx)/(x l nx)` equal to ?A. `l n (ln x) + c`B. `l n x + c`C. `(l nx)^(2) + c`D. None of the above |
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Answer» Correct Answer - A Let `I = int (dx)/(x l nx)` Put `l n x = t rArr (1)/(x) = dt` `:. I = int (1)/(t) dt = l n t + c = l n (l nx) + c` where c is the constant of integration |
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