Answer
$$2\ln |\sqrt{x}| -2\ln |1+\sqrt{x}|+C$$
Work Step by Step
Given $$\int \frac{d x}{x+x \sqrt{x}}$$
Let
$$u^2 =x \ \ \to \ \ 2udu=dx $$
Then
\begin{aligned}
\int \frac{d x}{x+x \sqrt{x}}&=\int \frac{2ud u}{u^2+u^2 \sqrt{u^2}}\\
&=\int \frac{2ud u}{u^2+u^3 }\\
&=\int \frac{2du }{u^2+u }
\end{aligned}
Use partial fraction , since
\begin{aligned}
\frac{2}{u\left(u+1\right)}&=\frac{a_0}{u}+\frac{a_1}{u+1}\\
&=\frac{a_0\left(u+1\right)+a_1u}{u\left(u+1\right)}\\
2&=a_0\left(u+1\right)+a_1u
\end{aligned}
at $u=0\to a_0=2,\ \ \ u= -1\to a_1 = -2$, then
\begin{aligned}
\int \frac{d x}{x+x \sqrt{x}} &=\int \frac{2du }{u^2+u } \\
&=\int \frac{2}{u}du-\int\frac{2}{u+1}du\\
&=2\ln |u| -2\ln |1+u|+C\\
&=2\ln |\sqrt{x}| -2\ln |1+\sqrt{x}|+C
\end{aligned}
