You may start by converting the difference of squares to numerator into a product such that:

`(e^x)^2 - 1 = (e^x - 1)(e^x + 1)`

You may substitute t for `e^x+1` such that:

`e^x = t => e^x dx = dt => dx = (dt)/t`

Changing the variable yields:

`int (e^(2x) - 1)/(e^(2x)+1)dx = int (t^2 - 1)/(t^2+1)*(dt)/t`

Using the linearity of integrals yields:

`int (t^2 - 1)/(t^2+1)*(dt)/t = int t^2/(t(t^2+1)) dt - int 1/(t(t^2+1))dt`

You need to add and subtract 1 to numerator of `int t^2/(t(t^2+1)) dt` such that:

`int t^2/(t(t^2+1)) dt = int (t^2+1-1)/(t(t^2+1)) dt`

`int t^2/(t(t^2+1)) dt = int (t^2+1)/(t(t^2+1)) dt -int 1/(t(t^2+1))dt`

`int t^2/(t(t^2+1)) dt = int 1/t dt - int 1/(t(t^2+1))dt`

`int (t^2 - 1)/(t^2+1)*(dt)/t = int 1/t dt - 2int 1/(t(t^2+1))dt`

You may solve the integral `int 1/(t(t^2+1))dt` using partial fraction decomposition such that:

`1/(t(t^2+1)) = A/t + (Bt + C)/(t^2+1)`

`1 = At^2 + A + Bt^2 + ` `Ct`

`1 = t^2(A+B) + Ct + A`

Equating the coefficients of like powers yields:

`A+B = 0 => A = -B`

`C = 0`

`A = 1 => B = -1`

`1/(t(t^2+1)) = 1/t - 1/(t^2+1)`

Integrating both sides yields:

`int 1/(t(t^2+1)) dt= int 1/t dt- int 1/(t^2+1) dt`

`int 1/(t(t^2+1)) dt = ln |t| - arctan t + c`

`int (t^2 - 1)/(t^2+1)*(dt)/t = ln |t| - 2(ln |t| - arctan t)+ c`

`int (t^2 - 1)/(t^2+1)*(dt)/t = 2 arctan t - ln|t| + c`

Substituting back `e^x` for t yields:

`int (e^(2x) - 1)/(e^(2x)+1)dx = 2 arctan(e^x) - ln(e^x)+ c`

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

**Hence, evaluating the given integral yields `int (e^(2x) - 1)/(e^(2x)+1)dx = 2 arctan (e^x) - x + c.` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now