Proof-Verification:$\int x[3+\ln(1+x^2)]\arctan x{\rm d}x$. $$\begin{aligned}
&\int x[3+\ln(1+x^2)]\arctan x{\rm d}x\\
=&\int 3x\arctan x{\rm d}x+\int x\ln(1+x^2)\arctan x{\rm d}x\\
=&\int \arctan x{\rm d}\left(\frac{3x^2}{2}\right)+\int \ln(1+x^2)\arctan x{\rm d}\left(\frac{x^2}{2}\right)\\
=&\int \arctan x{\rm d}\left(\frac{3x^2}{2}\right)+\int \arctan x{\rm d}\left[\frac{1}{2}(1+x^2)\ln(1+x^2)-\frac{1}{2}x^2\right]\\
=&\int \arctan x {\rm d} \left[ \frac{3x^2}{2}+\frac{1}{2}(1+x^2)\ln(1+x^2)-\frac{1}{2}x^2 \right]\\
=&\int \arctan x {\rm d} \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]
\end{aligned}$$
Integrating by parts, we obtain
$$\begin{aligned}
&\int \arctan x {\rm d} \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{1}{1+x^2}\cdot\left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]{\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{x^2}{1+x^2}{\rm d}x-\frac{1}{2}\int \ln(1+x^2){\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{x^2}{1+x^2}{\rm d}x-\frac{1}{2}x\ln(1+x^2)+\int \frac{x^2}{1+x^2}{\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\frac{1}{2}x\ln(1+x^2)+C.\\
\end{aligned}$$
Please correct me if I'm wrong.
 A: We can verify your solution by simply finding its derivative. If $F(x)$ is the antiderivative of $f(x)$, then we have $F'(x)=f(x)$.
$$F'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan(x)\left(x^2+\frac{1}{2}\left(x^2+1\right)\ln\left(x^2+1\right)\right)-\frac{1}{2}x\ln\left(x^2+1\right)+C\right]$$
Let us find the derivative of the last two terms first. The derivative of $C$, the constant of integration, is obviously zero.
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[-\frac{1}{2}x\ln\left(x^2+1\right)+C\right]=-\frac{1}{2}\left(\frac{2x^2}{x^2+1}+\ln\left(x^2+1\right)\right)=-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}$$
Next, let us find the derivative of the remaining portion of $F(x)$. For simplicity and ease of calculation, allow for the functions $g(x)=\arctan(x)$ and $h(x)=x^2+\frac{1}{2}\left(x^2+1\right)\ln\left(x^2+1\right)$. Apply the product rule.
$$F'(x)=g(x)h'(x)+h(x)g'(x)-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}$$
First, let us find $h'(x)$. The derivative of $g(x)$ should come to mind quickly.
$$h'(x)=2x+\frac{1}{2}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2\ln\left(x^2+1\right)+\ln\left(x^2+1\right)\right]=2x+\frac{x^3+x}{x^2+1}+x\ln\left(x^2+1\right)$$
Factor $x$ from each term of $h'(x)$ above and simplify. Multiply this result with the function $g(x)$ and we find that $g(x)h'(x)=x\left(3+\ln\left(x^2+1\right)\right)\arctan(x)$. This is clearly the original integrand $f(x)$, but we are not quite finished yet. Let us find $h(x)g'(x)$ first. We are almost finished verifying your solution.
$$h(x)g'(x)=\frac{1}{x^2+1}\left(x^2+\frac{\left(x^2+1\right)\ln\left(x^2+1\right)}{2}\right)=\frac{x^2}{x^2+1}+\frac{\ln\left(x^2+1\right)}{2}$$
Return to $F'(x)$. At this moment, only insert $h(x)g'(x)$. This will simplify very nicely for us.
$$F'(x)=g(x)h'(x)+\frac{x^2}{x^2+1}+\frac{\ln\left(x^2+1\right)}{2}-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}=g(x)h'(x)$$
Finally, insert $g(x)h'(x)$. You have solved the problem correctly.
$$F'(x)=f(x)=x\left(3+\ln\left(x^2+1\right)\right)\arctan(x)$$
