Since
u
=
2x
2
2x+5
and
dx
=
1
2(2x
1)
du,
substituting in the integral yields
2x
1
(2x
2
2x+5)
3
dx
=
2x
1
u
3
1
2(2x
1)
du
=
1
2u
3
du
canceling the term (2x
1)
2x+5
1)2x+51)
1
2x+5)3