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)

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