Need help integrating $e^{8x}/(e^{16x}+36)$ I started off substituting $u = e^x$ so $du = e^x\, dx$, giving me 
$$\int \frac{u^8}{u^{16}+36}\,du.$$
So then I substituted $v = u^8$ and $dv = 8u^7du$, which gives me 
$$\int \frac{1}{v^2 + 36}\,dv.$$
At this point I am unsure what to do.
 A: It's nice to recognize that the denominator is $(e^{8x})^2 + 6^2$. This suggests trying to get the integral into the form $\int \frac{1}{u^2+a^2}\,du$, which can be solved by elementary means.
So, let's try that: setting $w=e^{8x}$, we have $dw = 8e^{8x}\,dx$, so the integral becomes
$$\int\frac{e^{8x}\,dx}{(e^{8x})^2 + 36} = \frac{1}{8}\int \frac{dw}{w^2+36}.$$
Now, factor out $6^2$ from the denominator, to get
$$\frac{1}{8}\int\frac{dw}{w^2 + 36} = \frac{1}{288}\int\frac{{dw}}{\left(\frac{w}{6}\right)^2 + 1}.$$
Now set $u=\frac{w}{6}$, so $6du = dw$. We have:
$$\frac{1}{288}\int\frac{dw}{\left(\frac{w}{6}\right)^2 + 1}= \frac{1}{48}\int\frac{du}{u^2+1}.$$
The last integral should be immediate (Hint: $\frac{1}{1+t^2}$ is the derivative of a standard [inverse] function, so its integral is direct). Then taking $u=\frac{w}{6} = \frac{1}{6}e^{8x}$ gets you the integral; you can always differentiate to check.
A: Ok. I shall start from the first. Put $t =e^{8x}$ So you have
\begin{align*}
\int \frac{e^{8x}}{e^{16x} + 36} \ \text{dx} & = \frac{1}{8}\int \frac{1}{t^{2} + 36} \ \text{dt} \\ &= \frac{1}{8}\int \frac{1}{36 \tan^{2}{\theta} + 36} \times 6 \sec^{2}{\theta} \ d{\theta} \ \Bigl[ \text{ Putting} \ t=6\tan{\theta}\Bigr]\\ &= \frac{6}{8} \int \frac{1}{36 \cdot (1+\tan^{2}{\theta})} \cdot \sec^{2}{\theta} \ d\theta
\end{align*}
I am sure you can proceed from here. The substitution which you needed was $\tan{\theta}$. Remember the following:


*

*Denominator of the form: $a^{2}- x^{2}$ substitute $x = a\sin{\theta}$ or $a \cos{\theta}$

*Denominator of the form : $a^{2} + x^{2}$ substitute $x = a\tan{\theta}$.
