<s>[INST] $\int \frac{2x}{9x^2+3}dx=?$ So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way: 
\begin{align}
& \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right) \\[10pt]
= {} & \frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-x\frac{\arctan(\sqrt3x)}{\sqrt{3}}+\frac{\arctan(\sqrt3x)}{\sqrt3}\right)=\frac{\arctan(\sqrt3x)}{\sqrt3}+C
\end{align}
But at the same time, the textbook says, that it is equal to this:
$$
\frac{1}{9}\log(3x^2+1)+C
$$
Did I do something wrong? It would be quite hard for me to believ that these to functions are the same. 
Thanks for the responses! 
 [/INST] $$\int \frac{2x}{9x^2+3}dx$$
$$2\int \frac{x}{9x^2+3}dx$$
Apply Integral Substitution $u=9x^2+3\quad \:du=18xdx$
$$2\int \frac{1}{18u}du = 2\frac{1}{18}\int \frac{1}{u}du=2\frac{1}{18}\ln \left|u\right|=2\frac{1}{18}\ln \left|9x^2+3\right|$$
So
$$\int \frac{2x}{9x^2+3}dx=\color{red}{\frac{1}{9}\ln \left|9x^2+3\right|+C}$$
</s>