<s>[INST] Indefinite integral of a rational function: $\int\frac{6x+4}{x^2+4}\,dx$ Find $\displaystyle\int\frac{6x+4}{x^2+4}\,dx$
The question asks to find integral of the expression so I divided them into two parts: 
$$
\int\frac{6x}{x^2+4}\,dx
$$
and
$$\int\frac{4}{x^2+4}.
$$
So, for the first integral, I set $u=x^2+4$ and got $\int\frac{3}{u}\,du$ which is $3\ln|x^2+4|+c$. But I don't know how to integrate the second part. 
 [/INST] Since $\displaystyle \frac{d}{dx} \tan^{-1} x = \frac{1}{x^2+1}$,
$$
\frac{1}{2}\frac{d}{dx} \tan^{-1} (x/2) = \frac{1}{2} \cdot \frac{1}{1+(x/2)^2} \cdot \frac{1}{2} = \frac{1}{4+x^2}
$$
</s>