Observe that the denominator of the integrand takes the form of
\(a^2 + x^2\text{,}\) where
\(a = 1\text{.}\) Then trig sub may work out here.
Let \(x = \tan(\theta)\text{.}\) Then
\begin{equation*}
1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta)
\end{equation*}
and
\begin{equation*}
\tan^{-1}(x) = \tan^{-1}\left(\tan(\theta)\right) = \theta
\end{equation*}
In addition, we know that \(\dfrac{dx}{d\theta} = \sec^2(\theta)\text{,}\) which implies that \(dx = \sec^2(\theta)\, d\theta\text{.}\)
Now putting all the pieces together, we obtain
\begin{align*}
\int \frac{\tan^{-1}(x)}{1 + x^2}\, dx \amp=\int \frac{\theta}{\sec^2(\theta)}\cdot \sec^2(\theta)\, d\theta \\
\amp= \int \theta\, d\theta \\
\amp= \frac{\theta^2}{2} + C
\end{align*}
By our assumption of \(x = \tan(\theta)\text{,}\) we know that \(\theta = \tan^{-1}(x)\text{.}\) Hence, we obtain
\begin{equation*}
\frac{\theta^2}{2} + C = \frac{\left(\tan^{-1}(x)\right)^2}{2} + C
\end{equation*}
Solution. Alternative method (if you are super observant)
The integral can be rewritten as follows
\begin{equation*}
\int \frac{\tan^{-1}(x)}{1 + x^2}\, dx = \int \tan^{-1}(x) \cdot \frac{1}{1 + x^2}\, dx
\end{equation*}
Observe that \(\dfrac{1}{1 + x^2}\) is the derivative of \(\tan^{-1}(x)\text{,}\) which implies that a \(u\)-sub may work out here.
Let \(u = \tan^{-1}(x)\text{.}\) Then
\begin{equation*}
\frac{du}{dx} = \frac{1}{1 + x^2}
\qquad\implies\qquad
du = \frac{1}{1 + x^2}\, dx
\end{equation*}
We obtain
\begin{align*}
\int \tan^{-1}(x) \cdot \frac{1}{1 + x^2}\, dx \amp= \int u\, du \\
\amp= \frac{u^2}{2} + C \\
\amp= \frac{\left(\tan^{-1}(x)\right)^2}{2} + C
\end{align*}
