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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ dx x ^ 3 (1 + x ^ 6 )(2)/(3) = xf(x)(1 + x ^ 6 )(1)/(3) + C ,...

If ∫dxx3(1+x6)23=xf(x)(1+x6)13+C∫ \dfrac{dx}{{x}^{3}\left(1 + {x}^{6}\right)\dfrac{2}{3}} = xf\left(x\right)\left(1 + {x}^{6}\right)\dfrac{1}{3} + C, where CC is a constant of integration, then the function f(x)f\left(x\right) is equal to

Options

  1. A3x2\dfrac{3}{{x}^{2}}
  2. B−12x3- \dfrac{1}{2{x}^{3}}
  3. C−16x3- \dfrac{1}{6{x}^{3}}
  4. D−12x2- \dfrac{1}{2{x}^{2}}

Correct answer

B. −12x3- \dfrac{1}{2{x}^{3}}

Step-by-step solution

We have, ∫dxx3(1+x6)23=xf(x)(1+x6)13+C∫ \dfrac{dx}{{x}^{3}\left(1 + {x}^{6}\right)\dfrac{2}{3}} = xf\left(x\right)\left(1 + {x}^{6}\right)\dfrac{1}{3} + C

Now, I=∫dxx7(1x6+1)23I = ∫ \dfrac{dx}{{x}^{7}\left(\dfrac{1}{{x}^{6}} + 1\right)\dfrac{2}{3}}

Put, t=1x6+1t = \dfrac{1}{{x}^{6}} + 1⇒dt=−6x7dx⇒ dt = - \dfrac{6}{{x}^{7}}dx

∴I=−16∫dtt23=−12t13+C∴ I = - \dfrac{1}{6} ∫ \dfrac{dt}{t\dfrac{2}{3}} = - \dfrac{1}{2}t\dfrac{1}{3} + C, where CC is the constant of integration=−12(1x6+1)13+C=−12(1+x6)13x2+C= - \dfrac{1}{2}\left(\dfrac{1}{{x}^{6}} + 1\right)\dfrac{1}{3} + C = - \dfrac{1}{2}\dfrac{\left(1 + {x}^{6}\right)\dfrac{1}{3}}{{x}^{2}} + C

∴f(x)=−12x3∴ f\left(x\right) = - \dfrac{1}{2{x}^{3}}

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