Answer
$$ \int \frac{d x}{x^{2}+x \sqrt{x}}=-2 \ln \sqrt{x}-\frac{2}{\sqrt{x}}+2 \ln (\sqrt{x}+1)+C$$
Work Step by Step
Given $$ \int \frac{d x}{x^{2}+x \sqrt{x}}$$ \begin{array}{l}{\text { Let } u=\sqrt{x}, \text { so } u^{2}=x \text { and } 2 u d u=d x \text { . Then } }\\{I=\int \frac{d x}{x^{2}+x \sqrt{x}}=\int \frac{2 u d u}{u^{4}+u^{3}}=\int \frac{2 d u}{u^{3}+u^{2}}=\int \frac{2 d u}{u^{2}(u+1)}} \\ \text {As we have} \\ {\begin{aligned}&\frac{2}{u^{2}(u+1)}=\frac{A}{u}+\frac{B}{u^{2}}+\frac{C}{u+1} \\ \text {So, we get } \\ &\Rightarrow 2=A u(u+1)+B(u+1)+C u^{2}\end{aligned}\\ {}\\ \text { Setting } u=0 \text { gives } B=2 . \text { Setting } u=-1} \\ {\text { gives } C=2 . \text { Equating coefficients of } u^{2}, \text { we get } 0=A+C, \text { so } A=-2.}{ \\\text { Thus, }} \\ {\begin{aligned}I&=\int \frac{2 d u}{u^{2}(u+1)}\\ &=\int\left(\frac{-2}{u}+\frac{2}{u^{2}}+\frac{2}{u+1}\right) d u\\ &=-2 \ln |u|-\frac{2}{u}+2 \ln |u+1|+C\\ &=-2 \ln \sqrt{x}-\frac{2}{\sqrt{x}}+2 \ln (\sqrt{x}+1)+C\end{aligned}}\end{array}
