Question

How do you use partial fractions to find the integral `int (e^x)/((e^(2x)+1)(e^x-1))dx`?

Answer 1

The integral equals `-1/2arctan(e^x) - 1/2ln((e^x + 1)/|e^x- 1|)+ C`

Explanation:

First of all, the integral can be rewritten as

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

Now let `u = e^x`. Then `du = e^xdx` and `dx = (du)/e^x`.

`int e^x/((u^2+ 1)(u - 1)) * (du)/e^x`

`int 1/((u^2 + 1)(u - 1))du`

Now use partial fractions.

`(Au + B)/(u^2 + 1) + C/(u - 1) = 1/((u^2 + 1)(u - 1))`

`(Au + B)(u - 1) + C(u^2+ 1) = 1`

`Au^2 + Bu - B - Au + Cu^2 + C = 1`

`(A + C)u^2 + (B - A)u + (C - B) = 1`

We now write a system of equations:

`{(A + C = 0), (B - A = 0), (C - B = 1):}`

We can simplify to `C = -A` and `B = A` and substitute into the third equation.

`-A - A = 1`

`-2A = 1`

`A = -1/2`

This implies that `C = 1/2` and `B = -1/2`.

Our integral becomes:

`int(-1/2u - 1/2)/(u^2 + 1) + (1/2)/(u - 1)du`

`int (-1/2(u + 1))/(u^2 + 1) + 1/(2(u - 1))du`

`int -1/2(u + 1)/(u^2 + 1) + 1/(2(u - 1)) du`

`-1/2int (u + 1)/(u^2 + 1)du+ int 1/(2(u - 1))du`

`-1/2int u/(u^2 + 1) + 1/(u^2 + 1) du + int 1/(2(u - 1))du`

`-1/2arctanu - 1/2ln(u^2 + 1) + 1/2ln|u - 1| +C`

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

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

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

Practice Exercises

Solve the following integrals using partial fractions. You may use a u-substitution to begin.

a) `int (4x + 1)/(4x^2 + 12x + 9)dx`

b) `int sinx/(cosx(sinx - 1))dx`

Solutions

a) `ln|2x + 3| + 5/(4x + 6) + C`
b) `1/4ln|(sinx + 1)/(1 - sinx)| + 1/(2(sinx - 1)) + C`

Hopefully this helps!