Question

If $I=\int \frac{dx}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}=A{\left(\frac{x-1}{x+2}\right)}^{1/4}+C$, then $A$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is D $\frac{4}{3}$

$I=\int \frac{dx}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}=\int \frac{dx}{{(x-1)}^2\sqrt[4]{4{\left(\frac{x+2}{x-1}\right)}^5}}$
Substitute $\frac{x+2}{x-1}=t\Rightarrow \frac{-3}{{(x-1)}^2}dx=dt$
$I=-\frac{1}{3}\int \frac{dt}{t^{\frac{5}{4}}}=\frac{4}{3}{t}^{-\frac{1}{4}}=\frac{4}{3}\sqrt[4]{4\frac{x-1}{x+2}}+c$

Hence comparing the above expression with the one given in the question, we get $A=\frac{4}{3}$