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

3 Answers
Apr 12, 2018

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#

Apr 12, 2018

# 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.!

Apr 12, 2018

# 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!