Question

The integral $\int \frac{dx}{\left(x+1\right)^{\frac{3}{4}}\left(x-2\right)^{\frac{5}{4}}}$ is equal to :

• A. $4\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}} +C$
• B. $4\left(\frac{x-2}{x+1}\right)^{\frac{1}{4}} +C$
• C. $- \frac{4}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}} +C$
• D. $- \frac{4}{3}\left(\frac{x-2}{x+1}\right)^{\frac{1}{4}} +C$

Answer 1

Answer: (C)

$\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$
$=\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{2}(x-2)^{\frac{5}{4}2}}$
$=\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{2}(x-2)^{-3 / 4}}$
$=\int \frac{(x+1)^{\frac{-3}{4}}}{(x+2)^{\frac{-3}{4}}} \cdot \frac{1}{(x-2)^{2}} d x$
$=\frac{1}{-3} \int\left[\frac{x+1}{x-2}\right]^{-3 / 4} \cdot\left(\frac{-3}{(x-2)^{2}}\right) d x$
$=\frac{-1}{3} \frac{\left(\frac{x+1}{x-2}\right)^{\frac{3}{4}+1}}{\left(\frac{-3}{4}+1\right)}+C$
$=-\frac{4}{3}\left[\frac{x+1}{x-2}\right]^{1 / 4}+ C$