Question

`int2/(2x^2+2x)dx`?

Answer 1

`ln(abs(x/(x+1)))+C`

Explanation:

First we factor out 2:
`int1/(x^2+x)dx`

Then factorise the denominator:
`int1/(x(x+1))dx`

We need to split this into partial fractions:
`1=A(x+1)+Bx`

Using `x=0` gives us:
`A=1`

Then using `x=-1` gives us:
`1=-B`

Using this we get:
`int1/x-1/(x+1)dx`

`int1/xdx-int/(x+1)dx`

`ln(abs(x))-ln(abs(x+1_)+C`

`ln(abs(x/(x+1)))+C`