Question

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

Answer 1

`2ln|2x-1|-2ln|x|+1/x+C`

Explanation:

We need to find `A,B,C` such that

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

for all `x`.

Multiply both sides by `x^2(2x-1)` to get

`1=Ax(2x-1)+B(2x-1)+Cx^2`

`1=2Ax^2-Ax+2Bx-B+Cx^2`

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

Equating coefficients give us

`{(2A+C=0),(2B-A=0),(-B=1):}`

And thus we have `A=-2,B=-1,C=4`. Substituting this in the initial equation, we get

`1/(x^2(2x-1))=4/(2x-1)-2/x-1/x^2`

Now, integrate it term by term
`int\ 4/(2x-1)\ dx-int\ 2/x\ dx-int\ 1/x^2\ dx`

to get

`2ln|2x-1|-2ln|x|+1/x+C`

Answer 2

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

Explanation:

Perform the decomposition into partial fractions

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

`=(A(2x-1)+Bx(2x-1)+C(x^2))/(x^2(2x-1))`

The denominators are the same, compare the numerators

`1=A(2x-1)+Bx(2x-1)+C(x^2)`

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

Let `x=1/2`, `=>`, `1=C/4`, `=>`, `C=4`

Coefficients of `x^2`

`0=2B+C`

`B=-C/2=-4/2=-2`

Therefore,

`1/(x^2(2x-1))=-1/x^2-2/x+4/(2x-1)`

So,

`int(1dx)/(x^2(2x-1))=-int(1dx)/x^2-int(2dx)/x+int(4dx)/(2x-1)`

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