Chú ý rằng $\cos (\tan^{-1} x)= \frac{1}{\sqrt{1+x^2}}$. Để chứng minh cái này, đặt $\tan^{-1}x=a$ thì $\tan a=x$. Khi đó ta có $\cos (\tan^{-1}x)= \cos a = \frac{1}{\sqrt{1+(\tan a)^2}}= \frac{1}{\sqrt{1+x^2}}$.
Do đó $f(x)= \left( \cos (\tan^{-1}x) \right)^4= \frac{1}{(1+x^2)^2}=(1+x^2)^{-2}$.
Ta có
$$\int \frac{1}{(1+x^2)^2} dx = \int \frac{1}{1+x^2} dx - \int \frac{x^2}{(1+x^2)^2} dx= \tan^{-1}x - \int \frac{x^2}{(1+x^2)^2} dx$$
Đặt $u=x, v'=\frac{x}{(1+x^2)^2}$ thì $v= \frac{-1}{2(1+x^2)}$ nên
$$\begin{align*} \int \frac{x^2}{(1+x^2)^2} dx &= \int uv' = uv- \int u'v, \\
& = \frac{-x}{2(1+x^2)}- \int \frac{-1}{2(1+x^2)}dx, \\
& = \frac{-x}{2(1+x^2)}+ \frac 12 \tan^{-1}x.
\end{align*}$$
Do đó $\int f(x) dx= \int \frac{1}{(1+x^2)^2}dx = \frac 12 \tan^{-1}x + \frac{x}{2(x^2+1)}$.