`int (x^3 - 6x - 20)/(x + 5) dx` Find the indefinite integral.

Tuesday, March 31, 2015

`int(x^3-6x-20)/(x+5)dx`

Let's evaluate the integral by applying integral substitution,

Let u=x+5, `=>x=u-5`

du=dx

`int(x^3-6x-20)/(x+5)dx=int((u-5)^3-6(u-5)-20)/udu`

`=int((u^3-5^3-3u^2*5+3u*5^2)-6u+30-20)/udu`

`=int(u^3-125-15u^2+75u-6u+10)/udu`

`=int(u^3-15u^2+69u-115)/udu`

`=int(u^2-15u+69-115/u)du`

Now apply the sum rule,

`=intu^2du-int15udu-int115/udu+int69du`

`=intu^2du-int15udu-115int(du)/u+69intdu`

Use the following common integrals,

`intx^ndx=x^(n+1)/(n+1)`

and `int1/xdx=ln(|x|)`

`=u^3/3-15u^2/2-115ln|u|+69u`

Substitute back u=x+5,

`=(x+5)^3/3-15/2(x+5)^2-115ln|x+5|+69(x+5)`

Add a constant C to the solution,

`=(x+5)^3/3-15/2(x+5)^2+69(x+5)-115ln|x+5|+C`

Ninenotes