JEE PYQ: Indefinite Integration Question 20
Question 20 - 2019 (08 Apr Shift 2)
If $\int \frac{dx}{x^3(1+x^6)^{2/3}} = xf(x)(1+x^6)^{1/3} + C$, where $C$ is a constant of integration, then the function $f(x)$ is equal to:
(1) $\frac{3}{x^2}$
(2) $-\frac{1}{6x^3}$
(3) $-\frac{1}{2x^2}$
(4) $-\frac{1}{2x^3}$
Show Answer
Answer: (4)
Solution
$\int \frac{dx}{x^3(1+x^6)^{2/3}} = \int \frac{x^{-7},dx}{(x^{-6}+1)^{2/3}}$. Let $x^{-6}+1 = t$, $-6x^{-7},dx = dt$. $I = -\frac{1}{6}\int t^{-2/3},dt = -\frac{1}{2}t^{1/3} + C = -\frac{1}{2x^2}(1+x^6)^{1/3} + C$. Comparing: $xf(x) = -\frac{1}{2x^2}$, so $f(x) = -\frac{1}{2x^3}$.
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