Question

$\int \frac{1}{\left[\right(x-1{)}^3(x+2{)}^5{]}^{1/4}}dx$ is equal to

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

Answer 1

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

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