How to solve: $\int\frac{\mathrm{e}^x}{\mathrm{e}^{2x}+3}\,\mathrm{d}x$

Question

$${\displaystyle\int}\dfrac{\mathrm{e}^x}{\mathrm{e}^{2x}+3}\,\mathrm{d}x$$

I tried substituting $u= e^x$, $du = e^x dx$, and so I end up with:

$${\displaystyle\int}\dfrac{1}{\mathrm{u}^{2}+3}\,\mathrm{d}u$$

This looks like the derivative $\arctan(x)$, but I have a $3$ instead of a $1$. How can I solve this integral?

Answer 1

Substitute $u=\sqrt{3}t$. $${\displaystyle\int}\dfrac{1}{\mathrm{u}^{2}+3}\,\mathrm{d}u=\frac{1}{\sqrt{3}}{\displaystyle\int}\dfrac{1}{t^2+1}\,\mathrm{d}t=\frac{1}{\sqrt{3}}\arctan \left(t\right)=\frac{1}{\sqrt{3}}\arctan \left(\frac{u}{\sqrt{3}}\right)$$ Done! Now put back $u=e^x$ to get the answer.

Answer 2

You can write: $$\int \frac{1}{u^2+3}\ du= \int \frac{1}{3\left(\frac{u^2}{3}+1\right)}\ du= \frac{1}{3}\int \frac{1}{\left(\frac{u}{\sqrt3}\right)^2+1}\ du=\frac{1}{\sqrt3}\int \frac{\frac1{\sqrt3}}{\left(\frac{u}{\sqrt3}\right)^2+1}\ du= \\ =\frac{1}{\sqrt3}\arctan \left(\frac{u}{\sqrt{3}}\right)+C$$

without using a substitution.

Answer 3

$\int \frac{1}{u^2+3}\ du= \frac{1}{3}\int \frac{1}{\left(\frac{u}{\sqrt3}\right)^2+1}\ du=\frac{1}{\sqrt3}\tan^{-1} \left(\frac{u}{\sqrt{3}}\right)+C$