Question

How do you integrate `int (x+3)/(x^2(x-1) )` using partial fractions?

Answer 1

`4ln|(x-1)/x|+3/x+C, or, ln(1-1/x)^4+3/x+C`.

Explanation:

Here is a way to find the Integral `I=int(x+3)/{x^2(x-1)}dx`,

without using partial fractions.

`I=int(x+3)/{x^2(x-1)}dx`,

`=int{x/{x^2(x-1)}+3/{x^2(x-1)}}dx`,

`=int[1/{x(x-1)}+3*(x-(x-1)}/{x^2(x-1)}]dx`,

`=int[1/{x(x-1)}+3{x/{x^2(x-1)}-(x-1)/{x^2(x-1)}}]dx`,

`=int[1/{x(x-1)}+3/{x(x-1)}-3/x^2]dx`,

`=int4/{x(x-1)}dx-3int1/x^2dx`,

`=4int{x-(x-1)}/{x(x-1)}dx--3*x^(-2+1)/(-2+1)`,

`=4int{1/(x-1)-1/x}dx+3/x`,

`=4{ln|(x-1)|-ln|x|}+3/x`.

`rArr I=4ln|(x-1)/x|+3/x+C, or, ln(1-1/x)^4+3/x+C`.

Enjoy Maths.!