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

2 Answers
NickTheTurtle
Apr 17, 2018

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

Narad T.
Apr 17, 2018

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

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
11731 views around the world
You can reuse this answer
Creative Commons License