JEE PYQ: Indefinite Integration Question 31
Question 31 - 2019 (11 Jan Shift 1)
If $\int \frac{\sqrt{1-x^2}}{x^4},dx = A(x)\left(\sqrt{1-x^2}\right)^m + C$, for a suitable chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(A(x))^m$ equals:
(1) $\frac{-1}{27x^9}$
(2) $\frac{-1}{3x^3}$
(3) $\frac{1}{27x^6}$
(4) $\frac{1}{9x^4}$
Show Answer
Answer: (1)
Solution
Let $\frac{1}{x^2} - 1 = u^2$, then $-\frac{2}{x^3},dx = 2u,du$. $I = \int \frac{u}{x^4}\cdot(-x^3 u),du = -\int \frac{u^2}{x},du$. After computation: $A(x) = -\frac{1}{3x^3}$, $m = 3$. So $(A(x))^m = \left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}$.
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