Question

The integral $\int \frac{1}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}dx$ is equal to:
(where $C$ is a constant of integration)

• A. $\frac{3}{4}{\left(\frac{x+2}{x-1}\right)}^{\frac{5}{4}}+C$
• B. $\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{\frac{5}{4}}+C$
• C. $\frac{3}{4}{\left(\frac{x+2}{x-1}\right)}^{\frac{1}{4}}+C$
• D. $\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{\frac{1}{4}}+C$

Answer 1

The correct option is D $\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{\frac{1}{4}}+C$

$I=\int \frac{1}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}dx=\int \frac{1}{\sqrt[4]{4\frac{(x-1{)}^3}{(x+2{)}^3}(x+2{)}^8}}dx=\int \frac{dx}{(x+2{)}^2{\left(\frac{x-1}{x+2}\right)}^{3/4}}$
Let $\frac{x-1}{x+2}=t\Rightarrow \frac{3dx}{(x+2{)}^2}=dt$
$\therefore I=\int \frac{1}{t^{3/4}}\cdot \frac{1}{3}dt$
$=\frac{1}{3}\left(\frac{t^{-3/4+1}}{-\frac{3}{4}+1}\right)+C$
$=\frac{4}{3}\left(t^{1/4}\right)+C$
$\therefore I=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{\frac{1}{4}}+C$