Integrating e^-2x.tanh(x) - Solving for the Antiderivative

[DryRun]

Homework Statement

Find $$\int e^{-2x}\tanh x\,dx$$

The attempt at a solution

$$\int e^{-2x}\tanh x\,dx \\=\int e^{-2x}\times \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx \\=\int \frac{e^{-x}-e^{-3x}}{e^x+e^{-x}}\,dx$$

Then i have no idea how to proceed.

 

[RK7]

Hint: can you do anything with partial fractions?

 

[DryRun]

I've already tried to split it:
$$\int \frac{e^{-x}}{e^x+e^{-x}}-\int \frac{e^{-3x}}{e^x+e^{-x}}$$But got stuck again.

The answer is: $$\frac{e^{-2x}}{2}-\ln (1+e^{2x})$$

I'm thinking maybe integration by parts?

 

[SammyS]

I've already tried to split it:
$$\int \frac{e^{-x}}{e^x+e^{-x}}-\int \frac{e^{-3x}}{e^x+e^{-x}}$$But got stuck again.

The answer is: $$\frac{e^{-2x}}{2}-\ln (1+e^{2x})$$

I'm thinking maybe integration by parts?

$\displaystyle \frac{e^{-x}}{e^x+e^{-x}}=\frac{e^{-2x}}{1+e^{-2x}}$

$\displaystyle \frac{e^{-3x}}{e^x+e^{-x}}=\frac{e^{-4x}}{1+e^{-2x}}\ .$ Then let u = e^-x or maybe let u = 1+e^-2x .

 

[DryRun]

$$\int \frac{e^{-2x}}{1+e^{-2x}}\,.dx-\int \frac{e^{-4x}}{1+e^{-2x}}\,.dx$$
Let $u =e^{-x} \\Then \, \frac{du}{dx}=-e^{-x}=-u$
$$=\int \frac{-u}{1+u^2}\,.du-\int \frac{-u^3}{1+u^2}\,.du$$
$$=-\frac{1}{2}\ln (1+u^2) + \int u\,.du - \int \frac{u}{(1+u^2)}\,.du$$
$$=-\frac{1}{2}\ln (1+u^2) + \frac{u^2}{2} - \frac{1}{2}\ln (1+u^2)$$
$$=\frac{u^2}{2}-\ln (1+u^2)$$
Substituting for u, we have the final answer:
$$=\frac{e^{-2x}}{2}-\ln (1+e^{-2x})$$
But the answer in my copybook is:
$$=\frac{e^{-2x}}{2}-\ln (1+e^{2x})$$
By the way, isn't there a need to add the arbitrary constant of integration?

 

[ehild]

Your result is correct, and add a constant.

ehild

 

[DryRun]

Hi ehild!

Thank you for your confirmation.