How do you integrate `int (5x²-2x-1)/( (x+1)(x²+1))` using partial fractions?

Split the fraction up. `x+1` is linear, so its numerator will just be `A`, whereas `x^2+1` is irreducible (over the real numbers) so it will be in the form `Bx+C`.

`(5x^2-2x-1)/((x+1)(x^2+1))=A/(x+1)+(Bx+C)/(x^2+1)`

Which becomes

`5x^2-2x-1=A(x^2+1)+(Bx+C)(x+1)`

`5x^2-2x-1=Ax^2+A+Bx^2+Bx+Cx+C`

`5x^2-2x-1=x^2(A+B)+x(B+C)+A+C`

Giving the system:

`{(A+B=5),(B+C=-2),(A+C=-1):}`

Subtract the last from the middle to see that `B-A=-1`, which can be added to the first equation to see that `2B=4=>B=2`.

Substitute this to see that `A=3` and `C=-4`.

This gives the partial fraction decomposition of

`(5x^2-2x-1)/((x+1)(x^2+1))=3/(x+1)+(2x-4)/(x^2+1)`

So, we now want to find

`int3/(x-1)dx+int(2x-4)/(x^2+1)dx`

The first integral is simple:

`=3lnabs(x-1)+int(2x-4)/(x^2+1)dx`

In the second, you should realize the `(du)/u` pattern in `(2x)/(x^2+1)`, which is indicative of a natural log. However, this doesn't account for the leftover `-4`.

`=3lnabs(x-1)+int(2x)/(x^2+1)dx-int4/(x^2+1)dx`

We can now deal with the `(du)/u` we identified:

`=3lnabs(x-1)+ln(x^2+1)-int4/(x^2+1)dx`

Recall that the absolute value signs aren't necessary since `x^2+1` is always positive.

To deal with `4/(x^2+1)`, recognize that this is simply `4` times the derivative of the `arctan(x)` function, giving a final answer of

`=3lnabs(x-1)+ln(x^2+1)-4arctan(x)+C`