Helping Im stuck with Question of Integral of $\frac{e^{2x}}{6e^{x}+2}$

Question

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 $(\frac{e^x}{6})-(\frac{1}{18})ln|6e^x+2|+C$

Answer 1

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.

Answer 2

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}