<s>[INST] Is my calculation of the integral $\int \tan^{-1} x \, dx$ correct? Compute $\int \tan^{-1}x \,dx$.
First, set $u = \arctan(x)$ and $dv = dx$. We want to find $du$ and we already have $v = x$.
We start by taking the tangent of both sides, leaving us with
$$\tan(u) = x.$$
Next, using implicit differentiation, we get $\frac{du}{dx}\sec^2(u) = 1$, or $$\frac{du}{dx} = \frac{1}{\sec^2(u)} = \frac{1}{1+ \tan^2(u)} = \frac{1}{1+ \tan^2(\arctan(x))} = \frac{1}{x^2+1}.$$
Therefore, $du = \frac{1}{x^2 +1}\, dx.$
We know that $\int u\, dv = uv - \int v\, du$ by Integration By Parts.
So, $\int \arctan(x)\, dx = x\arctan(x) - \int \frac{x}{x^2+1}\, dx.$
We can simplify $\int \frac{x}{x^2+1}\, dx$ by using u-substitution, making a different $u = x^2+1$ and $\frac{du}{2x} = dx.$
Subbing these in, we get
$$\int \frac{x}{u} \cdot \frac{du}{2x} = \frac{1}{2}\int \frac{1}{u}\, du = \frac{1}{2}|\log u| = \frac{1}{2}|\log(x^2+1)| + C.$$
Plugging this back into $x\arctan(x) - \int \frac{x}{x^2+1}\, dx,$ we get
$\boxed{x\arctan(x) - \frac{1}{2} |\log(x^2+1)| + C}$, where C is a constant.
 [/INST] Yes, but you can simply do like below
$$\int \arctan x~\mathrm{d}x= x\cdot \arctan x - \int \frac{x}{x^2+1}\mathrm{d}x =  x\cdot \arctan x - \frac{1}{2}\log (x^2+1) +C$$
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