# Solving the integral by substitution

I am given an integral:

$$\int \frac{e^{4x}}{36+e^{8x}}\,\mathrm dx$$

I am told to solve this by substitution where $u=e^{4x}/6$, and that I need to write the integrand as a function of $u$.

I am just very lost on how to make the appropriate substitution. I have been able to complete previous questions, it is just that for this particular question, I am finding it very difficult to see how I can substitute $u$.

I have found that $\mathrm du = e^{4x} / 24\mathrm dx$

I would appreciate it if you could help me substitute $u$, I am just finding this part of the question difficult.

Thank you very much.

$$I=\int{\frac{e^{4x}}{36+e^{8x}}}\;dx$$

$$I=\frac{1}{36}\int{\frac{e^{4x}}{1+\frac{e^{8x}}{36}}}\; dx$$

$$I=\frac{1}{36}\int{\frac{e^{4x}}{1+\left(\frac{e^{4x}}{6}\right)^2}}\; dx$$

$$let\ u=\frac{e^{4x}}{6}\quad then\quad du=\frac{2}{3}e^{4x}dx\quad and\ replace\ in\ I$$

$$I=\frac{1}{36}\int \frac{\frac{3}{2}du}{1+u^2}$$

$$I=\frac{1}{36}\cdot\frac{3}{2}\int \frac{du}{1+u^2}$$

$$I=\frac{1}{24}\int \frac{du}{1+u^2}$$

$$I=\frac{1}{24}\arctan(u)\ +\ C$$ $$Back\ to\ x$$ $$\mathbf{I=\frac{1}{24}\arctan\left(\frac{e^{4x}}{6}\right)\ +\ C}$$

\begin{align} u &= e^{4x}/6 \\ du &= 4 e^{4x} / 6 \, dx \\ \frac{3}{2} \, du &= e^{4x} \, dx \\ \frac{3}{2} (???) \, du &= \frac{e^{4x}}{36 + e^{8x}} \, dx \end{align}