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$\begin{aligned} & \int(1+x) \log \left(1+x^2\right) d x=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \\ & \log \left(1+x^2\right)+g(x)+C, \text { then } g(x)=\end{aligned}$
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The correct answer is:
$-2 x-\frac{x^2}{2}+2 \tan ^{-1} x$
Here $\int(1+x) \log \left(1+x^2\right) d x$
$=\int \log \left(1+x^2\right) \cdot 1 d x+\int x \cdot \log \left(1+x^2\right) d x$
$\begin{aligned}=\log \left(1+x^2\right) \cdot x-\int\left(\frac{1}{1+x^2}\right) \cdot( & (2 x) \cdot x d x \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
$=x \log \left(1+x^2\right)-2 \int\left(\frac{x^2+1-1}{1+x^2}\right) d x+\int x \cdot \log \left(1+x^2\right) d x$
$\begin{aligned}=x \log \left(1+x^2\right)-2\left[\int\left(1-\frac{1}{1+x^2}\right) d x\right] & \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
$\begin{aligned} & =x \cdot \log \left(1+x^2\right)-2\left[\int 1 d x-\int\left(\frac{1}{1+x^2}\right) d x\right]+\int x \cdot \log \\ & =x \cdot \log \left(1+x^2\right)-2\left[x-\tan ^{-1} x\right]+\int x \cdot \log \left(1+x^2\right) d x \\ & =x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
Now from $\int \mathrm{x} \cdot \log \left(1+\mathrm{x}^2\right) \mathrm{dx}$
let $1+\mathrm{x}^2=\mathrm{t}$
$2 x d x=d t$
$\begin{aligned} & =\frac{1}{2} \int \log \mathrm{t} d \mathrm{t}=\frac{1}{2}\left[\log \mathrm{t} \cdot \mathrm{t}-\int\left(\frac{\mathrm{t}}{\mathrm{t}} \cdot \mathrm{t}\right) \mathrm{dt}\right] \\ & =-[\mathrm{t} \operatorname{lot} \mathrm{t} \quad \mathrm{t}] \\ & =\frac{1}{2}\left[\left(1+\mathrm{x}^2\right) \log \left(1+\mathrm{x}^2\right)-\left(1+\mathrm{x}^2\right)\right]\end{aligned}$
Now from given integration
$\begin{gathered}\int(H x) \log \left(1+x^2\right) d x=x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \\ +\frac{1}{2}\left[\left(1+x^2\right) \log \left(1+x^2\right)-\left(1+x^2\right)\right]+c_1\end{gathered}$
$\begin{array}{r}=\left(\mathrm{x}+\frac{\mathrm{x}^2}{2}+\frac{1}{2}\right) \log \left(1+\mathrm{x}^2\right)-2 \mathrm{x}+2 \tan ^{-1} \mathrm{x}+\frac{1}{2} \\ \left(1+\mathrm{x}^2\right)+\mathrm{c}_1\end{array}$
$=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$
$+\mathrm{C}, \mathrm{C}=-\frac{1}{2}+\mathrm{C}_1$
$\Rightarrow g(x)=-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$
$=\int \log \left(1+x^2\right) \cdot 1 d x+\int x \cdot \log \left(1+x^2\right) d x$
$\begin{aligned}=\log \left(1+x^2\right) \cdot x-\int\left(\frac{1}{1+x^2}\right) \cdot( & (2 x) \cdot x d x \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
$=x \log \left(1+x^2\right)-2 \int\left(\frac{x^2+1-1}{1+x^2}\right) d x+\int x \cdot \log \left(1+x^2\right) d x$
$\begin{aligned}=x \log \left(1+x^2\right)-2\left[\int\left(1-\frac{1}{1+x^2}\right) d x\right] & \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
$\begin{aligned} & =x \cdot \log \left(1+x^2\right)-2\left[\int 1 d x-\int\left(\frac{1}{1+x^2}\right) d x\right]+\int x \cdot \log \\ & =x \cdot \log \left(1+x^2\right)-2\left[x-\tan ^{-1} x\right]+\int x \cdot \log \left(1+x^2\right) d x \\ & =x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \int x \cdot \log \left(1+x^2\right) d x\end{aligned}$
Now from $\int \mathrm{x} \cdot \log \left(1+\mathrm{x}^2\right) \mathrm{dx}$
let $1+\mathrm{x}^2=\mathrm{t}$
$2 x d x=d t$
$\begin{aligned} & =\frac{1}{2} \int \log \mathrm{t} d \mathrm{t}=\frac{1}{2}\left[\log \mathrm{t} \cdot \mathrm{t}-\int\left(\frac{\mathrm{t}}{\mathrm{t}} \cdot \mathrm{t}\right) \mathrm{dt}\right] \\ & =-[\mathrm{t} \operatorname{lot} \mathrm{t} \quad \mathrm{t}] \\ & =\frac{1}{2}\left[\left(1+\mathrm{x}^2\right) \log \left(1+\mathrm{x}^2\right)-\left(1+\mathrm{x}^2\right)\right]\end{aligned}$
Now from given integration
$\begin{gathered}\int(H x) \log \left(1+x^2\right) d x=x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \\ +\frac{1}{2}\left[\left(1+x^2\right) \log \left(1+x^2\right)-\left(1+x^2\right)\right]+c_1\end{gathered}$
$\begin{array}{r}=\left(\mathrm{x}+\frac{\mathrm{x}^2}{2}+\frac{1}{2}\right) \log \left(1+\mathrm{x}^2\right)-2 \mathrm{x}+2 \tan ^{-1} \mathrm{x}+\frac{1}{2} \\ \left(1+\mathrm{x}^2\right)+\mathrm{c}_1\end{array}$
$=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$
$+\mathrm{C}, \mathrm{C}=-\frac{1}{2}+\mathrm{C}_1$
$\Rightarrow g(x)=-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$
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