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

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Answer 1 · 2017-01-06T21:28:10

See below.

#int3/((1+x)(1-2x))=ln(|1+x|)-ln(|1-2x|)+C#

The denominator is already factored.

#=>3/((1+x)(1-2x))=A/(1+x)+B/(1-2x)#

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

#(1+x)(1-2x)[3/(cancel((1+x)(1-2x)))=A/cancel(1+x)+B/cancel(1-2x)]#

=>#3=A(1-2x)+B(1+x)#

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

#x=1/2#

#3=cancel(A*0)+3/2B#

#B=2#

#x=-1#

#3=A(1-(-2))+cancel(B*0)#

#3=3A#

#A=1#

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

#int(1)/(1+x)+2/(1-2x)dx#

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

#int1/(1+x)dx+2int1/(1-2x)dx#

For the first integral, #u=1+x, du=dx#
For the second integral, #z=1-2x, dz=-2dx=>-1/2dz=dx#

#int1/udu-int1/zdz#

Integrate.

#ln(|u|)-ln(|z|)+C#

Substitute back in.

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

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

By the properties of logarithms, you may also write the final answer as:

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