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

Jun 21, 2016

$\int {e}^{2 x} / \left(3 + {e}^{2 x}\right) \setminus \mathrm{dx} = \ln \sqrt{3 + {e}^{2 x}} + C$

#### Explanation:

easiest way always is to recognise the patterns

generalisation $\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

so if we consider $\frac{d}{\mathrm{dx}} \ln \left(3 + {e}^{2 x}\right) = \frac{1}{3 + {e}^{2 x}} \cdot 2 {e}^{2 x}$ then we're pretty much done

because $\frac{d}{\mathrm{dx}} \ln \left(3 + {e}^{2 x}\right) = \frac{2 {e}^{2 x}}{3 + {e}^{2 x}}$ then we actually want $\frac{1}{2} \cdot \frac{d}{\mathrm{dx}} \ln \left(3 + {e}^{2 x}\right) = \frac{d}{\mathrm{dx}} \left[\frac{1}{2} \cdot \ln \left(3 + {e}^{2 x}\right)\right]$ moving the constant inside the derivative

$= \frac{d}{\mathrm{dx}} \left[\ln \sqrt{3 + {e}^{2 x}} \setminus\right]$

thusly

$\int {e}^{2 x} / \left(3 + {e}^{2 x}\right) \setminus \mathrm{dx} = \ln \sqrt{3 + {e}^{2 x}} + C$

you can plough through a whole series of subs but seeing the pattern is a real life saver.