JEE PYQ: Indefinite Integration Question 17
Question 17 - 2020 (09 Jan Shift 1)
The integral $\int \frac{dx}{(x+4)^{8/7}(x-3)^{6/7}}$ is equal to (where $C$ is a constant of integration):
(1) $\left(\frac{x-3}{x+4}\right)^{1/7} + C$
(2) $-\left(\frac{x-3}{x+4}\right)^{-1/7} + C$
(3) $\frac{1}{2}\left(\frac{x-3}{x+4}\right)^{3/7} + C$
(4) $-\frac{1}{13}\left(\frac{x-3}{x+4}\right)^{-13/7} + C$
Show Answer
Answer: (1)
Solution
Write as $\int \left(\frac{x-3}{x+4}\right)^{-6/7}\cdot\frac{1}{(x+4)^2},dx$. Let $\frac{x-3}{x+4} = t^7$. Then $\frac{7}{(x+4)^2},dx = 7t^6,dt$, so $\frac{dx}{(x+4)^2} = t^6,dt$. $I = \int t^{-6} \cdot t^6,dt = t + C = \left(\frac{x-3}{x+4}\right)^{1/7} + C$.
BETA
