How do you evaluate the integral #int dx/((2x-1)(x+2))#?

Question

5594 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T00:32:55

Use partial fractions. See below.

#int1/((2x-1)(x+2))dx=1/5ln(|2x-1|)-1/5ln(|x+2|)+C#

The denominator is already factored.

#=>1/((2x-1)(x+2))=A/(2x-1)+B/(x+2)#

Multiply through by the denominator of the left-hand side:

#(2x-1)(x+2)[1/cancel((2x-1)(x+2))=A/(cancel(2x-1))+B/cancel(x+2)]#

#=>1=A(x+2)+B(2x-1)#

We need to solve for #A# and #B#. One way to do this is to pick values for #x# which will cancel each variable.

#x=-2#

#1=B(-4-1)#

#1=-5B#

#B=-1/5#

#x=1/2#

#1=A(1/2+2)#

#1=A(5/2)#

#A=2/5#

We put these values back into our partial fractions and replace this as the integrand.

#=>int(2/5)/(2x-1)+(-1/5)/(x+2)dx#

Technically you should use a substitution before integrating. Split up the integral.

#=>2/5int1/(2x-1)dx-1/5int1/(x+2)dx#

For the first integral, #u=2x-1, du=2dx=> 1/2du=dx#.

For the second integral, #z=x+2, dz=dx#.

#=>1/5int1/udu-1/5int1/zdz#

Integrate.

#=>1/5ln|u|-1/5ln|z|+C#

Substitute back in.

#=>1/5ln|2x-1|-1/5ln|x+2|+C#

Note: the absolute value signs account for the domain of the natural log function (#x>0#).

By rules of logarithms, this is equivalent to

#1/5ln(|(2x-1)/(x+2)|)+C#