Answer
$\int \sqrt{x^{-\frac{2}{3}}+1} d x=\left(x^{\frac{2}{3}}+1\right)^{\frac{3}{2}}+C$
as $ u=1+x^{\frac{2}{3}}$
Work Step by Step
Given :
\[
\int \sqrt{x^{-\frac{2}{3}}+1} d x
\]
First, let us consider
\[
u=x^{\frac{2}{3}}+1
\]
$\Rightarrow d u=\frac{2}{3} x^{-\frac{1}{3}} d x$
\[
\Rightarrow \frac{3}{2} d u=x^{-\frac{1}{3}} d x
\]
Putting these values in the given integral, we have
\[
\begin{aligned}
&\int \sqrt{\frac{1}{x^{2}}+1} d x=\int \sqrt{x^{-\frac{2}{3}}+1} d x \\
&=\int \sqrt{\frac{x^{1+\frac{2}{3}}}{x^{\frac{2}{3}}}} d x \\
&=\int \frac{\sqrt{1+x^{\frac{2}{3}}}}{x^{\frac{1}{3}}} d x \\
&=\int[\sqrt{1+x^{\frac{2}{3}}}] x^{-\frac{1}{3}} d x \\
&=\int \frac{3}{2} u^{\frac{1}{2}} d u \quad \text { as } u=x^{\frac{2}{3}}+1 \\
&=\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right].\frac{3}{2}+C \\
&=\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right].\frac{3}{2}+C \\
&=u^{\frac{3}{2}}+C \\
&=\left(x^{1+\frac{2}{3}}\right)^{\frac{3}{2}}+C \quad \text { as } u=x^{\frac{2}{3}}+1
\end{aligned}
\]
