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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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451 Medium
451 of 995
Application +4 | -1
If ∫sin⁡  xsin⁡  (x−α)  dx=Ax+B  log⁡e  sin⁡ (x−α)+C\int \frac{{\sin \,\,x}}{{\sin \,\,\left( {x - \alpha } \right)}}\,\,dx = Ax + B\,\,{\log _e}\,\,\sin \,\left( {x - \alpha } \right) + C, then
View answer and explanation
Correct option: C
A=cos⁡  α,   B=sin⁡  αA = \cos \,\,\alpha ,\;\,B = \sin \,\,\alpha
Option C is correct because by expanding sin⁡x\sin x as sin⁡((x−α)+α)\sin((x-\alpha) + \alpha) and integrating, we find A=cos⁡αA = \cos \alpha and B=sin⁡αB = \sin \alpha.
452 Easy
452 of 995
Application +4 | -1
If ∫x3(1+x43)1/7dx=α(1+x43)β+c\int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{1 / 7} d x=\alpha\left(1+\sqrt[3]{x^4}\right)^\beta+c, then α+β−1\alpha+\beta^{-1} is
View answer and explanation
Correct option: D
4932\frac{49}{32}
Option D (4932\frac{49}{32}) is correct because after evaluating the integral and comparing coefficients, α+β−1\alpha+\beta^{-1} equals this value.
453 Medium
453 of 995
Application +4 | -1
∫x−6(1+2x3)2/3dx\int x^{-6}\left(1+2 x^3\right)^{2 / 3} d x equals
View answer and explanation
Correct option: A
−15(2+x−3)5/3+c\frac{-1}{5}\left(2+x^{-3}\right)^{5 / 3}+c
Option A (−15(2+x−3)5/3+c\frac{-1}{5}\left(2+x^{-3}\right)^{5 / 3}+c) is correct because after substitution and integration, the expression simplifies to this form.
454 Easy
454 of 995
Analyzing +4 | -1
∫dx(x  +  1)5  (x  + 2)34\int \frac{{dx}}{{\sqrt[4]{{{{\left( {x\,\, + \,\,1} \right)}^5}\,\,{{\left( {x\,\, + \,2} \right)}^3}}}}} is equal to
View answer and explanation
Correct option: C
− 4   (x  +  2x  +  1)1/4+  c- \,4\,\;{\left( {\frac{{x\,\, + \,\,2}}{{x\,\, + \,\,1}}} \right)^{1/4}} + \,\,c
Option C (− 4  (x + 2x + 1)1/4+ c- \,4\;{\left( {\frac{{x\, + \,2}}{{x\, + \,1}}} \right)^{1/4}} + \,c) is correct because it is the result of the integration after a suitable substitution.
455 Medium
455 of 995
Application +4 | -1
Let ∫  5  −  x2  +  x  dx\int \,\,\sqrt {\frac{{5\,\, - \,\,x}}{{2\,\, + \,\,x}}} \,\,dx equal
View answer and explanation
Correct option: B
x+2  5−x  +  7  sin⁡−1  x+27+C\sqrt {x + 2} \,\,\sqrt {5 - x} \,\, + \,\,7\,\,{\sin ^{ - 1}}\,\,\sqrt {\frac{{x + 2}}{7}} + C
Option B (x+2 5−x + 7 sin⁡−1 x+27 +C\sqrt {x + 2} \,\sqrt {5 - x} \, + \,7\,{\sin ^{ - 1}}\,\sqrt {\frac{{x + 2}}{7}} \, + C) is correct because after substituting 2+x=t22 + x = t^2, the integral can be solved using standard integral formulas.
456 Hard
456 of 995
Application +4 | -1
∫(xx)x(2xlog⁡ex+x)dx\int\left(x^x\right)^x\left(2 x \log _e x+x\right) d x is equal to
View answer and explanation
Correct option: B
(xx)x+C\left(x^x\right)^x+C
Option B is correct because the integral simplifies to the derivative of (x^x)^x, making the integration straightforward.
457 Easy
457 of 995
Application +4 | -1
∫axxdx=\int \frac{a^{\sqrt{x}}}{\sqrt{x}} d x=
View answer and explanation
Correct option: B
2axlog⁡ae+c2 a^{\sqrt{x}} \log _a e+c
Option B (2axlog⁡ae+c2 a^{\sqrt{{x}}} \log _a e+c) is correct because the integral can be solved by substitution and using the formula ∫axdx=axlog⁡ea+C=axlog⁡ae+C\int a^x dx = \frac{{a^x}}{{\log_e a}} + C = a^x \log_a e + C.
458 Medium
458 of 995
Application +4 | -1
The primitive function of the function f  (x)=  (a2−x2)x4f\,\,\left( x \right) = \,\,\frac{{\sqrt {\left( {{a^2} - {x^2}} \right)} }}{{{x^4}}} is
View answer and explanation
Correct option: C
c  −  (a2−x2)3/23a2 x3c\,\, - \,\,\frac{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}{{3{a^2}\,{x^3}}}
Option C (c,−,(a2−x2)3/23a2 x3c\\,\\ - \\,\frac{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}{{3{a^2}\,{x^3}}}) is correct because it is the result of integrating the given function using substitution.
459 Easy
459 of 995
Application +4 | -1
∫(1+x−x−1)ex+x−1dx=\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x=
View answer and explanation
Correct option: D
xex+x−1+Cx e^{x+x^{-1}}+C
Option D (xex+x−1+Cx e^{x+x^{-1}}+C) is correct because integrating by parts simplifies the expression, leading to the given result.
460 Easy
460 of 995
Application +4 | -1
∫  log⁡  (x  +  1+x2)1  +  x2    dx\int \,\,\frac{{\log \,\,\left( {x\,\, + \,\,\sqrt {1 + {x^2}} } \right)}}{{\sqrt {1\,\, + \,\,{x^2}} }}\;\,\,dx is equal to
View answer and explanation
Correct option: D
12  [log⁡  (x  +  1  +  x2)]2+  c\frac{1}{2}\,\,{\left[ {\log \,\,\left( {x\,\, + \,\,\sqrt {1\,\, + \,\,{x^2}} } \right)} \right]^2} + \,\,c
Option D is correct because the integral is solved using substitution, resulting in 12 [log⁡ (x + 1 +x2)]2 + c\frac{1}{2}\,{\left[ {\log \,\left( {x\,+\,\sqrt {1\,+{x^2}} } \right)} \right]^2}\,+\,c.
461 Medium
461 of 995
Analyzing +4 | -1
∫x2−1x2+1⋅dx1+x4\int \frac{x^2-1}{x^2+1} \cdot \frac{d x}{\sqrt{1+x^4}} is equal to
View answer and explanation
Correct option: D
12Cos⁡−1x2x2+1+c\frac{1}{\sqrt{2}} \operatorname{Cos}^{-1} \frac{x \sqrt{2}}{x^2+1}+c
Option D is correct because the integral simplifies to an inverse cosine form after dividing by x2 and using a substitution.
462 Easy
462 of 995
Application +4 | -1
If I=∫cos⁡2x−cos⁡2αsin⁡x−sin⁡αdxI=\int \frac{\cos 2 x-\cos 2 \alpha}{\sin x-\sin \alpha} d x, then II equals
View answer and explanation
Correct option: B
2cos⁡x−2xsin⁡α+C2 \cos x-2 x \sin \alpha+C
Option B (2cos⁡x−2xsin⁡α+C2 \cos x-2 x \sin \alpha+C) is correct because after simplifying the integral using trigonometric identities, the result matches this option.
463 Easy
463 of 995
Application +4 | -1
∫01  x71  −  x4    dx\int \nolimits_0^1 \,\,\frac{{{x^7}}}{{\sqrt {1\,\, - \,\,{x^4}} }}\;\,\,dx is equal to
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because it is the result of evaluating the definite integral using substitution and the Gamma function.
464 Medium
464 of 995
Application +4 | -1
∫xcos⁡x(xsin⁡x+cos⁡x)2dx\int \frac{x \cos x}{(x \sin x+\cos x)^2} d x is equal to
View answer and explanation
Correct option: B
−1(xsin⁡x+cos⁡x)+k-\frac{1}{(x \sin x+\cos x)}+k
Option B (−1(xsin⁡x+cos⁡x)+k\frac{-1}{(x \sin x+\cos x)}+k) is correct because the integral simplifies to this form after recognizing the derivative in the numerator.
465 Medium
465 of 995
Application +4 | -1
∫cos⁡(log⁡ex)dx\int \cos \left(\log _e x\right) d x is equal to
View answer and explanation
Correct option: C
x2[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+c\frac{x}{2}\left[\cos \left(\log _e x\right)+\sin \left(\log _e x\right)\right]+c
Option C (x2[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+c\frac{x}{2}\left[\cos \left(\log _e x\right)+\sin \left(\log _e x\right)\right]+c) is correct because by using integration by parts twice, we can solve the integral.
466 Easy
466 of 995
Understanding +4 | -1
Let f(x)=1xln⁡(xex)f(x)=\frac{1}{x} \ln \left(\frac{x}{e^x}\right) then its primitive w.r.t. xx is
View answer and explanation
Correct option: C
12ln⁡2x−x+C\frac{1}{2} \ln ^2 \mathrm{x}-\mathrm{x}+\mathrm{C}
Option C (12ln⁡2x−x+C\frac{1}{2} \ln ^2 \mathrm{x}-\mathrm{x}+\mathrm{C}) is correct because it is the result of integrating the given function after simplification using logarithm properties.
467 Easy
467 of 995
Analyzing +4 | -1
The value of I=∫dx(a+bx2)b−ax2I=\int \frac{d x}{\left(a+b x^2\right) \sqrt{b-a x^2}}, is
View answer and explanation
Correct option: A
1a(a2+b2)tan⁡−1(xa2+b2ab−ax2)+C\frac{1}{\sqrt{a\left(a^2+b^2\right)}} \tan ^{-1}\left(\frac{x \sqrt{a^2+b^2}}{a \sqrt{b-a x^2}}\right)+C
Option A is correct because the integral evaluates to 1a(a2+b2)tan⁡−1(xa2+b2ab−ax2)+C\frac{1}{\sqrt{a(a^2+b^2)}} \tan ^{-1}\left(\frac{x \sqrt{a^2+b^2}}{a \sqrt{b-a x^2}}\right)+C after using a trigonometric substitution and simplification.
468 Medium
468 of 995
Application +4 | -1
∫  x4+1(1−x4)3/2  dx\int \,\,\frac{{{x^4} + 1}}{{{{\left( {1 - {x^4}} \right)}^{3/2}}}}\,\,dx is equal to
View answer and explanation
Correct option: A
x1  −  x4  +  C\frac{x}{{\sqrt {1\,\, - \,\,{x^4}} }}\,\, + \,\,C
Option A (x1  −  x4  +  C\frac{x}{{\sqrt {1\,\, - \,\,{x^4}} }}\,\, + \,\,C) is correct because the integral can be solved by rewriting the expression and using a suitable substitution.
469 Medium
469 of 995
Application +4 | -1
∫  ∣ x ∣  log⁡  ∣ x ∣  dx\int \,\,\left| {\,x\,} \right|\,\,\log \,\,\left| {\,x\,} \right|\,\,dx equals (x≠0)\left( {x \ne 0} \right)
View answer and explanation
Correct option: D
12x  ∣ x ∣  log⁡   ∣ x ∣  −  14  x  ∣ x ∣  +  c\frac{1}{2}x\,\,\left| {\,x\,} \right|\,\,\log \,\,\,\left| {\,x\,} \right|\,\, - \,\,\frac{1}{4}\,\,x\,\,\left| {\,x\,} \right|\,\, + \,\,c
Option D (\frac{1}{2}x\,\left| {\,x\} \right|\,\log \,\,\left| {\,x\} \right|\, - \,\frac{1}{4}\,x\,\left| {\,x\} \right|\, + \,c\frac{1}{2}x\,\left| {\,x\} \right|\,\log \,\,\left| {\,x\} \right|\, - \,\frac{1}{4}\,x\,\left| {\,x\} \right|\, + \,c) is correct because it is the result of integrating |x|log|x| using integration by parts, considering both positive and negative cases for x.
470 Easy
470 of 995
Application +4 | -1
∫sec⁡xlog⁡(sec⁡x+tan⁡x)dx=\int \sec x \log (\sec x+\tan x) d x=
View answer and explanation
Correct option: B
12[log⁡(sec⁡x+tan⁡x)]2+c\frac{1}{2}[\log (\sec x+\tan x)]^2+c
Option B is correct because the integral evaluates to 12[log⁡(sec⁡x+tan⁡x)]2+c\frac{1}{2}[\log (\sec x+\tan x)]^2+c after a simple substitution.
471 Medium
471 of 995
Application +4 | -1
∫x2−1(x2+1)x4+1dx=\int \frac{x^2-1}{\left(x^2+1\right) \sqrt{x^4+1}} d x=
View answer and explanation
Correct option: B
12sec⁡−1(x2+12x)+c\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^2+1}{\sqrt{2} x}\right)+c
Option B (12sec⁡−1(x2+12x)+c\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^2+1}{\sqrt{2} x}\right)+c) is correct because it is the result of the integration after appropriate substitution and simplification.
472 Easy
472 of 995
Analyzing +4 | -1
If ∫cos⁡2xsin⁡4xcos⁡4x(1+cos⁡22x)dx=2log⁡e(1+cos⁡2x)−log⁡e(1+cos⁡22x)+f(x)+c\int \frac{\cos 2 x \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=2 \log _e(1+\cos 2 x)-\log _e\left(1+\cos ^2 2 x\right)+f(x)+\mathrm{c} then f(x)f(x) is
View answer and explanation
Correct option: A
sec⁡2x\sec ^2 x
Option A is correct because the integral can be solved by trigonometric manipulation and partial fractions, leading to the given expression for f(x).
473 Medium
473 of 995
Application +4 | -1
∫xa3−x3dx=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x=
View answer and explanation
Correct option: B
23Sin⁡−1(xa)3/2+c\frac{2}{3} \operatorname{Sin}^{-1}\left(\frac{x}{a}\right)^{3 / 2}+c
Option B (23Sin⁡−1(xa)3/2+c\frac{2}{3} \operatorname{Sin}^{-1}\left(\frac{x}{a}\right)^{3 / 2}+c) is correct because by substituting x=asin⁡2/3θx=a \sin ^{2 / 3} \theta, we can evaluate the integral.
474 Easy
474 of 995
Application +4 | -1
∫32x3(log⁡x)2dx\int 32 x^3(\log x)^2 d x is equal to
View answer and explanation
Correct option: A
x4{8(log⁡x)2−4(log⁡x)+1}+cx^4\left\{8(\log x)^2-4(\log x)+1\right\}+c
Option A (x4{8(log⁡x)2−4(log⁡x)+1}+cx^4\left\{8(\log x)^2-4(\log x)+1\right\}+c) is correct because it is the result of integrating the given expression using integration by parts twice.
475 Easy
475 of 995
Application +4 | -1
∫(x+1)(x+log⁡x)2xdx=\int \frac{(x+1)(x+\log x)^2}{x} d x=
View answer and explanation
Correct option: C
13(x+log⁡x)3+c\frac{1}{3}(x+\log x)^3+c
Option C is correct because the integral simplifies to 13(x+log⁡x)3+c\frac{1}{3}(x + \log x)^3 + c after applying the substitution method.
476 Easy
476 of 995
Application +4 | -1
∫  3x9x  −  1    dx\int \,\,\frac{{{3^x}}}{{\sqrt {{9^x}\,\, - \,\,1} }}\,\,\;dx is equal to
View answer and explanation
Correct option: A
1log⁡  3  log⁡  ∣ 3x  +  9x−1 ∣  +  c\frac{1}{{\log \,\,3}}\,\,\log \,\,\left| {\,{3^x}\,\, + \,\,\sqrt {{9^x} - 1\,} } \right|\,\, + \,\,c
Option A is correct because the integral can be solved by substituting 3x=t{3^x} = t and using the standard integral formula.
477 Medium
477 of 995
Application +4 | -1
∫1x2+2 d (x2+1)\int \frac{1}{\sqrt{x^2+2}} \,d\,\left(x^2+1\right) is equal to
View answer and explanation
Correct option: A
2x2+2+C2 \sqrt{x^2+2}+C
Option A (2x2+2+C2 \sqrt{{x^2+2}}+C) is correct because it is the direct result of integrating the given expression.
478 Medium
478 of 995
Analyzing +4 | -1
If ∫dx(x2+a2)3=x4a2(x2+a2)+mna2\int \frac{d x}{\left(x^2+a^2\right)^3}=\frac{x}{4 a^2\left(x^2+a^2\right)}+\frac{m}{n a^2} {x2a2(x2+a2)+12a3tan⁡−1(xa)}+C\left\{\frac{x}{2 a^2\left(x^2+a^2\right)}+\frac{1}{2 a^3} \tan ^{-1}\left(\frac{x}{a}\right)\right\}+C. Then ∣m−n∣|m-n| is equal to
View answer and explanation
Correct option: D
1
Option D is correct because after evaluating the integral and comparing coefficients, the absolute difference between m and n is found to be 1.
479 Easy
479 of 995
Application +4 | -1
∫ax1−a2xdx=\int \frac{a^x}{\sqrt{1-a^{2 x}}} d x=
View answer and explanation
Correct option: A
1log⁡asin⁡−1ax+c\frac{1}{\log a} \sin ^{-1} a^x+c
Option A is correct because the integral can be solved using substitution.
480 Easy
480 of 995
Application +4 | -1
∫  ex  (x−1)x2    dx\int \,\,{e^x}\,\,\frac{{\left( {x - 1} \right)}}{{{x^2}}}\;\,\,dx is equal to
View answer and explanation
Correct option: C
exx  +  c\frac{{{e^x}}}{x}\,\, + \,\,c
Option C (exx + c\frac{{{e^x}}}{x}\, + \,c) is correct because the integral can be simplified and solved using the property ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
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