How do you integrate int (x+3)/(x^2(x-1) )∫x+3x2(x−1) using partial fractions?

1 Answer
Ratnaker Mehta
Apr 15, 2018

4ln|(x-1)/x|+3/x+C, or, ln(1-1/x)^4+3/x+C4ln∣∣∣x−1x∣∣∣+3x+C,or,ln(1−1x)4+3x+C.

Explanation:

Here is a way to find the Integral I=int(x+3)/{x^2(x-1)}dxI=∫x+3x2(x−1)dx,

without using partial fractions.

I=int(x+3)/{x^2(x-1)}dxI=∫x+3x2(x−1)dx,

=int{x/{x^2(x-1)}+3/{x^2(x-1)}}dx=∫{xx2(x−1)+3x2(x−1)}dx,

=int[1/{x(x-1)}+3*(x-(x-1)}/{x^2(x-1)}]dx=∫[1x(x−1)+3⋅x−(x−1)x2(x−1)]dx,

=int[1/{x(x-1)}+3{x/{x^2(x-1)}-(x-1)/{x^2(x-1)}}]dx=∫[1x(x−1)+3{xx2(x−1)−x−1x2(x−1)}]dx,

=int[1/{x(x-1)}+3/{x(x-1)}-3/x^2]dx=∫[1x(x−1)+3x(x−1)−3x2]dx,

=int4/{x(x-1)}dx-3int1/x^2dx=∫4x(x−1)dx−3∫1x2dx,

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

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

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

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

Enjoy Maths.!

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
1331 views around the world
You can reuse this answer
Creative Commons License