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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Medium

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Showing 151-180 of 411 questions

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151 Medium
151 of 411
Application +4 | -1
If I=∫1∞dxx21+xI=\int_1^{\infty} \frac{d x}{x^2 \sqrt{1+x}}, then I equals
View answer and explanation
Correct option: C
log⁡(2+1)+2\log (\sqrt{2}+1)+\sqrt{2}
Option C is correct because the integral simplifies to the integral of -1 from -1 to 1, due to the odd function properties, resulting in -2.
152 Medium
152 of 411
Application +4 | -1
∫ex(x2+1)(x+1)2dx=\int e^x \frac{\left(x^2+1\right)}{(x+1)^2} d x=
View answer and explanation
Correct option: D
ex(x−1x+1)+ce^x\left(\frac{x-1}{x+1}\right)+c
Option D (ex(x−1x+1)+ce^x\left(\frac{x-1}{x+1}\right)+c) is correct because the integral is solved using the form ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
153 Medium
153 of 411
Application +4 | -1
∫sin⁡2xsin⁡4x+cos⁡4xdx=\int \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x=
View answer and explanation
Correct option: B
tan⁡−1(tan⁡2x)+c\tan ^{-1}\left(\tan ^2 x\right)+c
Option B is correct because the integral can be solved by rewriting the integrand using trigonometric identities and then applying substitution.
154 Medium
154 of 411
Application +4 | -1
If ∫dx1+tan⁡x=px+qlog⁡e∣cos⁡x+sin⁡x∣+c\int \frac{d x}{1+\tan x}=p x+q \log _e|\cos x+\sin x|+c then p+qp+q equals
View answer and explanation
Correct option: D
1
Option D (1) is correct because it is the sum of p and q, which are calculated using the formula for integrals of the form ∫dx1+tan⁡x\int \frac{{dx}}{{1 + \tan x}}.
155 Medium
155 of 411
Application +4 | -1
∫1{(x−1)3  (x+2)5}1/4dx\int \frac{1}{{{{\left\{ {{{\left( {x - 1} \right)}^3}\,\,{{\left( {x + 2} \right)}^5}} \right\}}^{1/4}}}}dx is equal to
View answer and explanation
Correct option: A
43  (x  −  1x  +  2)1/4+  C\frac{4}{3}\,\,{\left( {\frac{{x\,\, - \,\,1}}{{x\,\, + \,\,2}}} \right)^{1/4}} + \,\,C
Option A is correct because the integral simplifies to a form where the power rule can be applied after a suitable substitution.
156 Medium
156 of 411
Application +4 | -1
∫cot⁡−1(1−x+x2−x3+x4−x5+…)dx(∣x∣<1)\int \cot ^{-1}\left(1-x+x^2-x^3+x^4-x^5+\ldots\right) d x(|x| < 1)
View answer and explanation
Correct option: A
(x+1)tan⁡−1(x+1)−log⁡1+(x+1)2+C(x+1) \tan ^{-1}(x+1)-\log \sqrt{1+(x+1)^2}+C
Option A ((x+1)tan⁡−1(x+1)−log⁡1+(x+1)2+C(x+1) \tan ^{-1}(x+1)-\log \sqrt{1+(x+1)^2}+C) is correct because it is the result of integrating the given expression using integration by parts and simplifying the result.
157 Medium
157 of 411
Application +4 | -1
If u=∫eaxsin⁡bx dxu=\int e^{a x} \sin b x \,d x and v=∫eaxcos⁡bx dx,v=\int e^{a x} \cos b x \,d x, then (u2+v2)(a2+b2)=\left(u^2+v^2\right)\left(a^2+b^2\right)=
View answer and explanation
Correct option: B
e2axe^{2 a x}
Option B, e2axe^{2 a x}, is correct because when uu and vv are squared and added, and then multiplied by (a2+b2)(a^2+b^2), the result simplifies to e2axe^{2ax}.
158 Medium
158 of 411
Application +4 | -1
∫  x+sin⁡  x1+cos⁡  x  dx\int \,\,\frac{{x + \sin \,\,x}}{{1 + \cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: A
x  tan⁡  x2+Cx\,\,\tan \,\,\frac{x}{2} + C
Option A is correct because the integral simplifies to x  tan⁡  x2+Cx\,\,\tan \,\,\frac{x}{2} + C after applying integration by parts.
159 Medium
159 of 411
Application +4 | -1
∫cos⁡2x(sin⁡x+cos⁡x)2dx=\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x=
View answer and explanation
Correct option: A
log⁡∣sin⁡x+cos⁡x∣+c\log |\sin x+\cos x|+c
Option A (log⁡∣sin⁡x+cos⁡x∣+c\log |\sin x+\cos x|+c) is correct because the integral simplifies to this expression after using the trigonometric identity for cos⁡2x\cos 2x and simplifying the fraction.
160 Medium
160 of 411
Application +4 | -1
∫sec⁡2θ(sec⁡θ+tan⁡θ)2dθ\int \sec ^2 \theta(\sec \theta+\tan \theta)^2 d \theta
View answer and explanation
Correct option: C
(sec⁡θ+tan⁡θ)3[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\frac{(\sec \theta+\tan \theta)}{3}[2+\tan \theta(\sec \theta+\tan \theta)]+C
Option C ((sec⁡θ+tan⁡θ)3[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\frac{(\sec \theta+\tan \theta)}{3}[2+\tan \theta(\sec \theta+\tan \theta)]+C) is correct because after substituting t=sec⁡θ+tan⁡θt = \sec \theta + \tan \theta and simplifying, we obtain the given result.
161 Medium
161 of 411
Application +4 | -1
∫11+cos⁡2xdx=\int \frac{1}{1+\cos ^2 x} d x=
View answer and explanation
Correct option: C
12tan⁡−1(12tan⁡x)+c\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan x\right)+c
Option C is correct because the integral is solved by dividing by cos²x and using a substitution, resulting in 12tan⁡−1(12tan⁡x)+c\frac{1}{\sqrt{2}} \tan^{-1}(\frac{1}{\sqrt{2}} \tan x)+c.
162 Medium
162 of 411
Analyzing +4 | -1
If ∫f(x)cos⁡  x dx=12  {f(x)}2+  c,   then    f(x)\int f\left( x \right)\cos \,\,x\,dx = \frac{1}{2}\,\,{\left\{ {f\left( x \right)} \right\}^2} + \,\,c,\;\,\mathrm{then}\,\,\;f\left( x \right) is

View answer and explanation
Correct option: B
sin⁡  x+c\sin \,\,x + c
Option B (sin⁡  x+c\sin \,\,x + c) is correct because the derivative of 12{f(x)}2+c\frac{1}{2}\{f(x)\} ^2 + c leads to f′(x)=cos⁡xf'(x) = \cos x, and integrating gives f(x)=sin⁡x+cf(x) = \sin x + c.
163 Medium
163 of 411
Analyzing +4 | -1
If ∫  2x2+  3(x2−1)  (x2+4)  dx=a  log⁡  (x−1x+1)  +  b  tan⁡−1  (x2)+c\int \,\,\frac{{2{x^2} + \,\,3}}{{\left( {{x^2} - 1} \right)\,\,\left( {{x^2} + 4} \right)}}\,\,dx = a\,\,\log \,\,\left( {\frac{{x - 1}}{{x + 1}}} \right)\,\, + \,\,b\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{2}} \right) + c, then the value of aa and bb are
View answer and explanation
Correct option: D
(12,  12)\left( {\frac{1}{2},\,\,\frac{1}{2}} \right)
Option D (12,  12)\left( {\frac{{1}}{{2}},\,\,\frac{{1}}{{2}}} \right) is correct because by partial fraction decomposition and integration, we find that a=12a = \frac{1}{2} and b=12b = \frac{1}{2}.
164 Medium
164 of 411
Analyzing +4 | -1
∫(esin⁡xesin⁡x+x)2(1x−cos⁡x)dx=log⁡e(f(x))+g(x)+c\int\left(\frac{e^{\sin x}}{e^{\sin x}+x}\right)^2\left(\frac{1}{x}-\cos x\right) d x=\log _e(f(x))+g(x)+c; where cc is constant, then f(π)−g(π)f(\pi)-g(\pi) is
View answer and explanation
Correct option: B
π−1π+1\frac{\pi-1}{\pi+1}
Option B is correct because after the substitution and integration, the expression simplifies to the given form, and evaluating f(π) - g(π) yields (π-1)/(π+1).
165 Medium
165 of 411
Application +4 | -1
∫cos⁡3xelog⁡(sin⁡x)dx\int \cos ^3 x e^{\log (\sin x)} d x is equal to
View answer and explanation
Correct option: B
−cos⁡4x4+c-\frac{\cos ^4 x}{4}+c
Option B is correct because elog⁡(sin⁡x)=sin⁡xe^{\log(\sin x)} = \sin x, so the integral becomes ∫cos⁡3xsin⁡xdx\int \cos^3 x \sin x dx, which is solved by substituting t=cos⁡xt = \cos x.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
166 Medium
166 of 411
Application +4 | -1
If f(x)=x+tan⁡xf(x)=x+\tan x and ff is inverse of gg, then g′(x)g^{\prime}(x) is equal to
View answer and explanation
Correct option: C
12+(g(x)−x)2\frac{1}{2+(g(x)-x)^2}
Option C (12+(g(x)−x)2\frac{{1}}{{2+(g(x)-x)^2}}) is correct because it is the derivative of the inverse function g(x) found by differentiating the original function and using the relationship between a function and its inverse.
167 Medium
167 of 411
Application +4 | -1
Let A=[x2601−5120x]A=\left[\begin{array}{ccc}x^2 & 6 & 0 \\ 1 & -5 & 1 \\ 2 & 0 & x\end{array}\right] and B=[400010008]B=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8\end{array}\right]. If a function is defined as f(x)=tr⁡(AB)f(x)=\operatorname{tr}(A B), then ∫3dxf(x)\int \frac{3 d x}{f(x)} is equal to
View answer and explanation
Correct option: A
14ln∣2x−12x+5∣+C\frac{1}{4}ln\left| {\frac{{2x - 1}}{{2x + 5}}} \right| + C
Option A (14ln∣2x−12x+5∣+C\frac{1}{4}ln\left| {\frac{{2x - 1}}{{2x + 5}}} \right| + C) is correct because after calculating tr(AB) and integrating, the result matches this option.
168 Medium
168 of 411
Application +4 | -1
If f(3x−43x+4)=x+2f\left(\frac{3 x-4}{3 x+4}\right)=x+2 then ∫f(x)dx\int f(x) d x is equal to
View answer and explanation
Correct option: B
−83ln⁡∣(1−x)∣+23x+c-\frac{8}{3} \ln |(1-x)|+\frac{2}{3} x+c
Option B is correct because it represents the correct integral after substituting and simplifying the given function.
169 Medium
169 of 411
Analyzing +4 | -1
The value of ∫(x−1)(x+1)x3+x2+xdx\int \frac{(x-1)}{(x+1) \sqrt{x^3+x^2+x}} d x, is
View answer and explanation
Correct option: C
2tan⁡−1x2+x+1x+C2 \tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}+C
Option C (2tan⁡−1x2+x+1x+C2 \tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}+C) is correct because the integral simplifies to this form after appropriate substitutions and trigonometric manipulations.
170 Medium
170 of 411
Application +4 | -1
If ∫dx1+sin⁡x=tan⁡(x2+a)+c\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}+a\right)+c, then aa is equal to
View answer and explanation
Correct option: D
−π4-\frac{\pi}{4}
Option D (−π4-\frac{\pi}{4}) is correct because the integral evaluates to tan⁡(x2−π4)+c\tan(\frac{x}{2} - \frac{\pi}{4}) + c, thus a=−π4a = -\frac{\pi}{4}.
171 Medium
171 of 411
Analyzing +4 | -1
∫−11  17 x5  −  x4  +  29 x3−  31 x  +  1x2+  1  dx    is\int \nolimits_{ - 1}^1 \,\,\frac{{17\,{x^5}\,\, - \,\,{x^4}\,\, + \,\,29\,{x^3} - \,\,31\,x\,\, + \,\,1}}{{{x^2} + \,\,1}}\,\,dx\,\,\;\mathrm{is}
View answer and explanation
Correct option: C
43\frac{4}{3}
Option C (43\frac{4}{3}) is correct because the integral of the odd part is zero, and the integral of the even part is calculated to be 43\frac{4}{3}.
172 Medium
172 of 411
Application +4 | -1
∫e2x−1e2x+1dx=\int \frac{e^{2 x}-1}{e^{2 x}+1} d x=
View answer and explanation
Correct option: B
log⁡(e2x+1)−x+c\log \left(e^{2 x}+1\right)-x+c
Option B is correct because the integral can be solved by rewriting the integrand and using substitution.
173 Medium
173 of 411
Application +4 | -1
Primitive of f(x)=x⋅2ln⁡(x2+1)f(x)=x \cdot 2^{\ln \left(x^2+1\right)} w.r.t. xx is
View answer and explanation
Correct option: C
(x2+1)ln⁡2+12(ln⁡2+1)+C\frac{\left(\mathrm{x}^2+1\right)^{\ln 2+1}}{2(\ln 2+1)}+\mathrm{C}
Option C ((x2+1)ln⁡2+12(ln⁡2+1)+C\frac{\left(x^2+1\right)^{\ln 2+1}}{2(\ln 2+1)}+C) is correct because after substituting t=x2+1t = x^2 + 1 and integrating, we arrive at the given result.
174 Medium
174 of 411
Analyzing +4 | -1
∫ex2(2x+x3)(3+x2)2dx\int \frac{e^{x^2}\left(2 x+x^3\right)}{\left(3+x^2\right)^2} d x is equal to:
View answer and explanation
Correct option: D
12ex2(3+x2)+c\frac{1}{2} \frac{e^{x^2}}{\left(3+x^2\right)}+c
The correct answer is D (12ex2(3+x2)+c\frac{1}{2} \frac{e^{x^2}}{\left(3+x^2\right)}+c) because the integral can be solved by recognizing a pattern related to the derivative of ex2e^{x^2} and applying appropriate simplification.
175 Medium
175 of 411
Application +4 | -1
∫x+sin⁡x1+cos⁡xdx\int \frac{x+\sin x}{1+\cos x} d x is equal to
View answer and explanation
Correct option: A
xtan⁡x2+cx \tan \frac{x}{2}+c
Option A is correct because the integral simplifies to xtan⁡x2+cx \tan \frac{x}{2}+c after using trigonometric identities and integration by parts.
176 Medium
176 of 411
Application +4 | -1
∫x−3(sin⁡−1(ln⁡x)+cos⁡−1(ln⁡x))dx\int \sqrt{x-3}\left(\sin ^{-1}(\ln x)+\cos ^{-1}(\ln x)\right) d x is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the integrand is not defined for any real number, making the integral non-existent.
177 Medium
177 of 411
Analyzing +4 | -1
∫dxsin⁡x+sec⁡x\int \frac{d x}{\sin x+\sec x} is equal to 123log⁡e∣3+t3−t∣+Tan⁡−1z+c\frac{1}{2 \sqrt{3}} \log _e\left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+\operatorname{Tan}^{-1} z+c where
View answer and explanation
Correct option: B
t=sin⁡x−cos⁡x,z=sin⁡x+cos⁡xt=\sin x-\cos x, z=\sin x+\cos x
Option A (t=sin⁡x−cos⁡x,z=sin⁡x+cos⁡xt=\sin x-\cos x, z=\sin x+\cos x) is correct because the given substitutions correctly separate the integral into two standard forms.
178 Medium
178 of 411
Understanding +4 | -1
∫f′  (x)f(x)  log⁡  {f(x)}dx=\int \frac{{f'\,\,\left( x \right)}}{{f\left( x \right)\,\,\log \,\,\left\{ {f\left( x \right)} \right\}}}dx =
View answer and explanation
Correct option: C
log⁡  {log⁡  f(x) }  +  C\log \,\,\left\{ {\log \,\,f\left( x \right)\,} \right\}\,\, + \,\,C
Option C (\log \,\left\{ {\log \,f\left( x \right)\} \right\}\, + \,C\log \,\left\{ {\log \,f\left( x \right)\} \right\}\, + \,C) is correct because it is the result of applying u-substitution twice.
179 Medium
179 of 411
Understanding +4 | -1
Assertion: It is not possible to find ∫e−x2dx\int \mathrm{e}^{-\mathrm{x}^2} \mathrm{dx} by inspection method.

Reason : Function is not expressible in terms of elementary functions.
View answer and explanation
Correct option: A
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the integral of e(-x2) cannot be expressed in terms of elementary functions, making it impossible to find by inspection.
180 Medium
180 of 411
Application +4 | -1
If ∫  [log⁡ (log⁡  x)  +  1(log⁡  x)2]dx=x  [ f(x)  −  g(x) ]  +  c\int \,\,\left[ {\log \,(\log \,\,x)\,\, + \,\,\frac{1}{{{{\left( {\log \,\,x} \right)}^2}}}} \right]dx = x\,\,\left[ {\,f\left( x \right)\,\, - \,\,\mathrm{g}\left( x \right)\,} \right]\,\, + \,\,c, then
View answer and explanation
Correct option: A
f  (x)=log⁡ (log⁡  x);  g(x)=1log⁡  xf\,\,\left( x \right) = \log \,(\log \,\,x);\,\,\mathrm{g}\left( x \right) = \frac{1}{{\log \,\,x}}
Option A (f  (x)=log⁡ (log⁡  x);  g(x)=1log⁡  xf\,\,\left( x \right) = \log \,(\log \,\,x);\,\,\mathrm{g}\left( x \right) = \frac{1}{{\log \,\,x}}) is correct because after integrating by parts, the resulting expression matches the given form with these functions.
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