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

Question

Emily Ammerman · Accepted Answer

See below.

#int3/((1+x)(1-2x))=ln(|1+x|)-ln(|1-2x|)+C# It has already factored the denominator. #=>3/((1+x)(1-2x))=A/(1+x)+B/(1-2x)# Divide the result by the left-hand side's denominator: #(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 substitute these values as the integrand and add them back into our partial fractions. #int(1)/(1+x)+2/(1-2x)dx# In theory, you should divide the integral into smaller parts and use a substitution before integrating. #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# Assimilate. #ln(|u|)-ln(|z|)+C# Replace once more. #ln(|1+x|)-ln(|1-2x|)+C# Note: the absolute value signs account for the domain of the natural log function (#x>0#). You can also write the final solution as follows using the logarithm's properties: #ln(|(1+x)/(1-2x)|)+C#

Samantha Adkison · Answer

undefined To integrate $ \frac{3}{{(1 + x)(1 - 2x)}} $ using partial fractions, you first express the fraction in terms of partial fractions as follows:

$ \frac{3}{{(1 + x)(1 - 2x)}} = \frac{A}{1 + x} + \frac{B}{1 - 2x} $

Next, you find the values of A and B by multiplying both sides by $ (1 + x)(1 - 2x) $ and simplifying:

$ 3 = A(1 - 2x) + B(1 + x) $

Solve for A and B:

$ 3 = A - 2Ax + B + Bx $

Match coefficients of x terms and constant terms:

For x term: $ -2A + B = 0 $
For constant term: $ A + B = 3 $

Solve the system of equations to find A and B:

$ A = \frac{3}{3} = 1 $
$ B = 3 - A = 3 - 1 = 2 $

Now integrate each term separately:

$ \int \frac{1}{1 + x} \, dx = \ln|1 + x| + C_1 $
$ \int \frac{2}{1 - 2x} \, dx = -\frac{1}{2} \ln|1 - 2x| + C_2 $

Combine the results:

$ \int \frac{3}{{(1 + x)(1 - 2x)}} \, dx = \ln|1 + x| - \frac{1}{2} \ln|1 - 2x| + C $