# How to integrate int(e^2x+e^-2x)/(e^2x-e^-2x)dx?

May 23, 2018

We can rewrite as

$I = \int \frac{{e}^{2 x} + \frac{1}{e} ^ \left(2 x\right)}{{e}^{2 x} - \frac{1}{e} ^ \left(2 x\right)} \mathrm{dx}$

$I = \int \frac{\frac{{e}^{4 x} + 1}{e} ^ \left(2 x\right)}{\frac{{e}^{4 x} - 1}{e} ^ \left(2 x\right)} \mathrm{dx}$

$I = \int \frac{{e}^{4 x} + 1}{{e}^{4 x} - 1} \mathrm{dx}$

Let $u = 4 x$. Then $\mathrm{du} = 4 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{4}$

$I = \frac{1}{4} \int \frac{{e}^{u} + 1}{{e}^{u} - 1} \mathrm{du}$

Now use partial fractions.

$\frac{A}{{e}^{u} - 1} + \frac{B}{1} = \frac{{e}^{u} + 1}{{e}^{u} - 1}$

$A + B {e}^{u} - B = {e}^{u} + 1$

We can readily see that $B = 1$ and therefore $A = 2$.

Thus

$I = \frac{1}{4} \left(\int \frac{2}{{e}^{u} - 1} + 1 \mathrm{du}\right)$

$I = \frac{1}{2} \ln | {e}^{u} - 1 | + \frac{1}{4} u + C$

$I = \frac{1}{2} \ln | {e}^{4 x} - 1 | + x + C$

Hopefully this helps!

May 23, 2018

The answer is =1/2ln(1/2(|e^(2x)-e^-(2x)|)+C

#### Explanation:

The function is

$\frac{\left({e}^{2 x} - {e}^{-} \left(2 x\right)\right)}{\left({e}^{2 x} - {e}^{-} \left(2 x\right)\right)} = \coth \left(2 x\right) = \cosh \frac{2 x}{\sinh} \left(2 x\right)$

Therefore, the integral is

$I = \int \frac{\left({e}^{2 x} - {e}^{-} \left(2 x\right)\right) \mathrm{dx}}{{e}^{2 x} - {e}^{-} \left(2 x\right)}$

$= \int \coth \left(2 x\right) \mathrm{dx}$

$= \int \frac{\cosh \left(2 x\right) \mathrm{dx}}{\sinh} \left(2 x\right)$

Let $u = \sinh \left(2 x\right)$, $\implies$, $\mathrm{du} = 2 \cosh \left(2 x\right) \mathrm{dx}$

So, the integral is

$I = \frac{1}{2} \int \frac{\mathrm{du}}{u}$

$= \frac{1}{2} \ln \left(u\right)$

$= \frac{1}{2} \ln \left(\sinh \left(2 x\right)\right) + C$

=1/2ln(1/2(|e^(2x)-e^-(2x)|)+C#

May 23, 2018

We have,

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

Let's Substitute $u = {e}^{2 x} - {e}^{- 2 x}$.

So, $\mathrm{du} = 2 {e}^{2 x} - \left(- 2 {e}^{- 2 x}\right) \mathrm{dx} \Rightarrow \mathrm{du} = 2 {e}^{2 x} + 2 {e}^{- 2 x} \mathrm{dx}$

$\Rightarrow \mathrm{dx} = \frac{1}{2 {e}^{2 x} + 2 {e}^{- 2 x}} \mathrm{du}$

So,

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

$= \int \frac{\cancel{{e}^{2 x} + {e}^{- 2 x}}}{u} \times \left(\frac{1}{2 \cancel{\left({e}^{2 x} + {e}^{- 2 x}\right)}}\right) \mathrm{du}$

$= \frac{1}{2} \int \frac{1}{u} \mathrm{du}$

$= \frac{1}{2} \ln | u | + C$

$= \frac{1}{2} \ln | {e}^{2 x} - {e}^{- 2 x} | + C$

$= \frac{1}{2} \ln | {e}^{- 2 x} \left({e}^{4 x} - 1\right) | + C$

$= \frac{1}{2} \ln | {e}^{- 2 x} | + \frac{1}{2} \ln | {e}^{4 x} - 1 | + C$

$= \frac{1}{2} \times - 2 x + \frac{1}{2} \ln | {e}^{4 x} - 1 | + C$

$= \frac{1}{2} \ln | {e}^{4 x} - 1 | - x + C$

Hope this helps.