Question

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

Answer 1

The answer is `=ln(|x|)+2ln (|x-1|)+C`

Explanation:

Perform the decomposition into partial fractions

`(3x-1)/(x^2-x)=(3x-1)/(x(x-1))=A/(x)+B/(x-1)`

`=(A(x-1)+Bx)/(x(x-1))`

The denominators are the same, compare the numerators

`3x-1=A(x-1)+Bx`

Let `x=0`, `=>`, `-1=-A`, `=>`, `A=1`

Let `x=1`, `=>`, `2=B`, `=>`, `B=2`

Therefore,

`(3x-1)/(x^2-x)=1/(x)+2/(x-1)`

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

`=ln(|x|)+2ln (|x-1|)+C`

Answer 2

`ln|x|+2ln|(x-1)|+C, or, ln|x(x-1)^2|+C`.

Explanation:

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

To decompose the integrand `(3x-1)/{x(x-1)}` into partial

fraction, we let, for some `A,B in RR`,

`(3x-1)/{x(x-1)}=A/x+B/(x-1)={A(x-1)+Bx}/{x(x-1)}`,

`rArr A(x-1)+Bx=3x-1...........(diamond)`.

`"Note that "(diamond)" must hold "AA x in RR," in particular so, for"`

`x=0, and x=1`.

`x=0, x=1, and (diamond) rArr A=1, &, B=2," resp."`.

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

`=ln|x|+2ln|(x-1)|`.

`rArr I=ln|x|+2ln|(x-1)|+C, or, ln|x(x-1)^2|+C`.

Enjoy Maths.!

Answer 3

`2ln|(x-1)|+ln|x|+C`.

Explanation:

The Question can easily be solved without using the

Method of Partial Fraction, as shown below :

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

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

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

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

`=3intdx/(x-1)-intdx/(x-1)+intdx/x`,

`=2intdx/(x-1)+intdx/x`,

`=2ln|(x-1)|+ln|x|+C`, as before!

Enjoy Maths!