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<br /> \\=\int e^{-2x}\times \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx<br /> \\=\int \frac{e^{-x}-e^{-3x}}{e^x+e^{-x}}\,dx<br />

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}<br /> \\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.