Question

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

Answer 1

We can rewrite as

`I = int (e^(2x) + 1/e^(2x))/(e^(2x) - 1/e^(2x)) dx`

`I = int((e^(4x) + 1)/e^(2x))/((e^(4x) - 1)/e^(2x)) dx`

`I = int(e^(4x) + 1)/(e^(4x) - 1)dx`

Let `u = 4x`. Then `du = 4dx` and `dx= (du)/4`

`I = 1/4 int (e^u + 1)/(e^u - 1)du`

Now use partial fractions.

`A/(e^u - 1) + B/1 = (e^u + 1)/(e^u - 1)`

`A + Be^u - B = e^u + 1`

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

Thus

`I = 1/4(int 2/(e^u - 1) + 1 du)`

`I = 1/2ln|e^u - 1| + 1/4u + C`

`I = 1/2ln|e^(4x) - 1| + x + C`

Hopefully this helps!

Answer 2

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

Explanation:

The function is

`((e^(2x)-e^-(2x)))/((e^(2x)-e^-(2x)))=coth(2x)=cosh(2x)/sinh(2x)`

Therefore, the integral is

`I=int((e^(2x)-e^-(2x))dx)/(e^(2x)-e^-(2x))`

`=intcoth(2x)dx`

`=int(cosh(2x)dx)/sinh(2x)`

Let `u=sinh(2x)`, `=>`, `du=2cosh(2x)dx`

So, the integral is

`I=1/2int(du)/(u)`

`=1/2ln(u)`

`=1/2ln(sinh(2x))+C`

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

Answer 3

We have,

`int (e^(2x) + e^(-2x))/(e^(2x) - e^(-2x))dx`

Let's Substitute `u = e^(2x) - e^(-2x)`.

So, `du = 2e^(2x) - (-2e^(-2x))dx rArr du = 2e^(2x) + 2e^(-2x)dx`

`rArr dx = 1/(2e^(2x) + 2e^(-2x)) du`

So,

`int (e^(2x) + e^(-2x))/(e^(2x) - e^(-2x))dx`

`= int cancel(e^(2x) + e^(-2x))/u xx (1/(2cancel((e^(2x) + e^(-2x))))) du`

`= 1/2 int 1/(u) du`

`= 1/2 ln|u| + C`

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

`= 1/2 ln|e^(-2x)(e^(4x) - 1)| + C`

`= 1/2 ln |e^(-2x)| + 1/2 ln|e^(4x) - 1| + C`

`= 1/2 xx -2x + 1/2 ln|e^(4x) - 1| + C`

`= 1/2 ln|e^(4x) - 1| - x + C`

Hope this helps.