# How do you find the antiderivative of  (e^(2x))(sqrt(1+3e^(2x)))?

Aug 1, 2016

$= \frac{1}{9} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{3}{2}} + C$

#### Explanation:

there's a simple pattern here

$\frac{d}{\mathrm{dx}} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{3}{2}}$

$= \frac{3}{2} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{1}{2}} \cdot 6 {e}^{2 x}$

$= 9 {e}^{2 x} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{1}{2}}$

so $\frac{d}{\mathrm{dx}} \left(\frac{1}{9} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{3}{2}}\right) = {e}^{2 x} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{1}{2}}$

so

$\int \setminus {e}^{2 x} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{1}{2}} \setminus \mathrm{dx} = \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{9} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{3}{2}}\right) \setminus \mathrm{dx}$

$= \frac{1}{9} {\left(1 + 3 {e}^{2 x}\right)}^{\frac{3}{2}} + C$

of course you might wish to use a sub $u = {e}^{2 x}$ etc, but this way allows you to do it pretty much in your head.