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

1 Answer
Apr 29, 2018

#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#