easy Integrate needs help >_<

∫ dx/(x³·√(9-4x²)
Let u = 9-4x², du=-8xdx
∫ dx/(x³·√(9-4x²)
=∫ [(u-9)/4][√(u)]·du/8
={1/8}∫{[u^(3/2)-9u^(1/2)]/4}du
={1/8}{[5u^(5/2)-27u^(3/2)]/8}+C where C is an arbitrary constant
=[1/64][5u^(5/2)-27u^(3/2)]+C

∫(2→0)[(x²+x+5)dx]/[(x+1)(x²+4)]
=∫(2→0)[(x+1+x²+4)dx]/[(x+1)(x²+4)]
=∫(2→0){[1/(x²+4)]+[(x²+4)/(x+1)]}dx
=∫(2→0)[1/(x²+4)]dx+∫(2→0)[(x-1)+5/(x+1)]dx
=∫(2→0)[1/(x²+4)]dx+∫(2→0)(x-1)dx+∫(2→0)[5/(x+1)]dx
=∫(2→0)[1/(x²+4)]dx+[(x²/2-x](2→0)+5∫(2→0)[1/(x+1)]dx
=∫(2→0)[1/(x²+4)]dx+5[ln|x+1|](2→0)
=∫(2→0)[1/(x²+4)]dx+5ln3
Let x=2tanu, dx=2sec²udu
When x=2, u=π/4
When x=0, u=0
∫(2→0)[1/(x²+4)]dx+5ln3
=∫(u=π/4→u=0){[2sec²u/(4tan²u+4)]du}+5ln3
=∫(u=π/4→u=0)[(2sec²u/4sec²u)du]+5ln3
=∫(u=π/4→u=0)[(1/2)du]+5ln3
=[u/2](u=π/4→u=0)+5ln3
=π/8+5ln3

回答者：cf

he used u = 9-4x²
I think it is not uesd by by Trigonometric Subtitution(eg. sin ,cos , tan...)
also =[1/64][5u^(5/2)-27u^(3/2)]+C is not the finally answer
u should be returned back to in terms of x