integrate: ((x^4-2x)/(x+4))?

““`_____x^3-4x²+16x-66

x-4)x^4+0x^3+0x-2x

“`- x^4+4x______

““““`_-4x^3+0x²

“““““`-4x^3-16x²______

“““““““““_16x²-2x

““““““““““16x²+64x_____

““““““““““““`_-66x+0

“““““““““““““`-66x-264

So…(x^4-2x)/(x+4) = x^3-4x²+16x-66 + 264/x+4

∫(x^3-4x²+16x-66)dx + ∫(264/x+4)dx –> u=x+4 du=dx

x^4/4-4x^3/3+8x²-66x + ∫(264/u)du

x^4/4-4x^3/3+8x²-66x + 264∫(1/u)du

x^4/4-4x^3/3+8x²-66x + 264ln|x+4| +C

Now ur problem becomes

{(y-4)^4 -2(y-4)}dy/(y)

expand the series, divide each term by y, and integrate separate term…

After you get the answer in terms of y, re-substitute y= x+4 to get the final answer

Hope that helps