Answer
$$ \frac{1}{3}x^3 +\frac{1}{2}x-\frac{1}{4}\sin2x+2x\sin x +2\cos x+C$$
Work Step by Step
\begin{aligned}
\int(x+\sin x)^{2} d x&=\int(x^2+2\sin x+\sin^2x) d x\\
&=\int(x^2+2x\sin x+\frac{1}{2}-\frac{1}{2}\cos2x) d x\\
&=\frac{1}{3}x^3 +\frac{1}{2}x-\frac{1}{4}\sin2x+2\int x\sin xdx
\end{aligned}
To evaluate $\int x\sin xdx $ , use integration by parts , since
\begin{aligned}
u&=x\ \ \ \ \ dv=\cos xdx\\
du&=dx\ \ \ \ \ v=\sin x
\end{aligned}
Then
\begin{aligned}
\int x\sin xdx&=x\sin x-\int \sin xdx \\
&=x\sin x +\cos x+c_1
\end{aligned}
Hence
\begin{aligned}
\int(x+\sin x)^{2} d x &=\frac{1}{3}x^3 +\frac{1}{2}x-\frac{1}{4}\sin2x+2\int x\sin xdx\\
&= \frac{1}{3}x^3 +\frac{1}{2}x-\frac{1}{4}\sin2x+2x\sin x +2\cos x+C
\end{aligned}
