integrate $\int \frac{dx}{e^{2x}+e^{x}+6}$

Question

$$\int \frac{dx}{e^{2x}+e^{x}+6}$$

I have tried to start with substitution.

Is $u=e^x$ than $du=e^xdx\Rightarrow \frac{du}{e^x}=dx$ therefore we have

$$\int \frac{du}{u(u^{2}+u+6)}$$

Is applicable?

Answer 1

You're correct. To continue, use Partial Fractions: $$\frac{1}{u\left(u^2+u+6\right)}=-\frac{1}{6}\cdot \frac{u+1}{u^2+u+6}+\frac{1}{6}\cdot \frac{1}{u}$$

$$\int \frac{du}{u\left(u^2+u+6\right)}=-\frac{1}{12}\left(\int \frac{d\left(u^2+u+6\right)}{u^2+u+6}+\int \frac{du}{u^2+u+6}\right)+\frac{1}{6}\int \frac{du}{u}$$

$$=-\frac{1}{12}\left(\ln\left|u^2+u+6\right|+2\int \frac{d(2u+1)}{(2u+1)^2+23}\right)+\frac{1}{6}\ln|u|$$

Let $2u+1=t\sqrt{23}$.

$$\int\frac{d(2u+1)}{(2u+1)^2+23}=\frac{1}{\sqrt{23}}\int \frac{dt}{t^2+1}=\frac{1}{\sqrt{23}}\arctan t+C$$