Question
Evalute : $\int \log \left(x^2+x\right) d x$

Answer

$
\begin{aligned}
& \text { Let } I=\int \log \left(x^2+x\right) d x=\int\left[\log \left(x^2+x\right)\right] \cdot 1 d x \\
= & {\left[\log \left(x^2+x\right)\right] \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(x^2+x\right)\right\} \cdot \int d x\right] d x } \\
= & {\left[\log \left(x^2+x\right)\right] \cdot x-\int \frac{1}{x^2+x} \cdot \frac{d}{d x}\left(x^2+x\right) \times x d x } \\
= & x \log \left(x^2+x\right)-\int \frac{1}{x(x+1)} \cdot(2 x+1) \cdot x d x \\
= & x \log \left(x^2+x\right)-\int \frac{2 x+1}{x+1} d x \\
= & x \log \left(x^2+x\right)-\int \frac{2(x+1)-1}{x+1} d x \\
= & x \log \left(x^2+x\right)-\int\left(2-\frac{1}{x+1}\right) d x \\
= & x \log \left(x^2+x\right)-2 \int 1 d x+\int \frac{1}{x+1} d x \\
= & x \log \left(x^2+x\right)-2 x+\log |x+1|+c .
\end{aligned}
$

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