Helping Im stuck with Question of Integral of $\frac{e^{2x}}{6e^{x}+2}$ So for this question I set $u= e^x$ and $du= e^x dx$, if put it back in it will be integral of $\frac{u}{6u+2}$. How would I continue on? Do I have to break it down into parts if so how?
P.S: the answer should come out to $(\frac{e^x}{6})-(\frac{1}{18})ln|6e^x+2|+C$
 A: Your substitution is fine and you end up with 
$$ \int \frac{u}{6u+2} \, du . $$
Now to deal with the new integrand just rewrite your rational function:
$$ \frac{u}{6u+2} = \frac16 - \frac13 \frac1{6u+2} $$
(by, for example, polynomial long division).  Then proceed from there.
A: If you do what Andre recommended. Let $u = 6e^x + 2$. In order to write the top, we can square $u$.
\begin{align}
u^2 = (6e^x + 2)^2 &= 36e^{2x} + 24e^x +4
\end{align}
So
\begin{equation}
e^{2x} = u^2 - 35e^{2x} - 24e^{x} - 4
\end{equation}
Also $\frac{du}{6e^x} = dx$. Now we can solve the integral(remember $6e^x = u - 2$)
\begin{align}
\int \frac{e^{2x}}{6e^x + 2} dx &= \int \frac{u^2 - 35e^{2x} - 24e^x - 4}{u(u - 2)} du\\
&= \int \frac{u}{u-2} - \frac{24e^x}{u(6e^x)} - \frac{4}{u(u - 2)} du - \int \frac{35e^{2x}}{u} \frac{du}{6e^x}\\
&= u - 2\ln|u| + C - 35\int \frac{e^{2x}}{6e^x  + 2} dx\\
36 \int \frac{e^{2x}}{6e^x  + 2} dx &= 6e^x + 2 + C - 2\ln|6e^x + 2|\\
\int \frac{e^{2x}}{6e^x  + 2} dx &=\frac{e^x}{6} -\frac{1}{18}\ln|6e^x + 2| + C
\end{align}
A: Continuing from your substitution, and as an alternative to long division, you can manipulate the numerator so that the fraction is in an easier form to integrate:
$$
\begin{align}
\frac{u}{6u+2} &= \frac16\frac{6u}{6u+2} \\
&= \frac16\left(\frac{6u+2-2}{6u+2}\right) \\
&= \frac16\left(\frac{6u+2}{6u+2}-\frac2{6u+2}\right) \\
&= \frac16 - \frac13\frac1{6u+2}
\end{align}$$
