Question

How do you integrate `int ((x-3)(x-1))/(-x(x-5)) dx` using partial fractions?

Answer 1

`I=-x+3/5ln|x|-8/5ln|x-5|+C`

Explanation:

Here,

`I=int((x-3)(x-1))/(-x(x-5))dx`

`=-int(x^2-4x+3)/(x^2-5x`

`=-int(x^2-5x+x+3)/(x^2-5x)`

`=-int(x^2-5x)/(x^2-5x)dx-int(x+3)/(x^2-5x)dx`

`=-int1dx-int(x+3)/(x(x-5))dx`

`I=-x-I_1...to(D)`

We try to obtain Partial Fractions for `I_1`

`(x+3)/(x(x-5))=A/x+B/(x-5)`

`x+3=A(x-5)+Bx`

`x=0=>3=A(-5)=>A=-3/5`

`x=5=>5+3=5B=>B=8/5`

So

`I_1=int(-3/5)/xdx+int(8/5)/(x-5)dx`

`I_1=-3/5int1/xdx+8/5int1/(x-5)dx`

`I_1=-3/5ln|x|+8/5ln|x-5|+c`

Thus, from `(D)`,we get

`I=-x-{-3/5ln|x|+8/5ln|x-5|}+C`

`I=-x+3/5ln|x|-8/5ln|x-5|+C`