How do you integrate #(2x+1 )/ ((x-2)(x^2+4))# using partial fractions?

Question

Charles Anctil · Accepted Answer

#5/8ln|x-2|+3/8tan^-1(x/2)-5/16ln|x^2+4|+C#

We want to find #A#, #B#, and #C# such that #(2x+1)/((x-2)(x^2+4))=A/(x-2)+(Bx+C)/(x^2+4)#.

We multiply both sides by #(x-2)(x^2+4)# to get #2x+1=A(x^2+4)+(Bx+C)(x-2)#. Since we want to find #A#, #B#, and #C# such that all #x# satisfies the above equation, we can arbitrarily substitute values that simplify the problem, such as #x=2#.

Then, the equation would become #5=8A#. #A# then is #5/8#.

We can also substitute #x=0#: #1=4A-2C#. However, since #A=5/8#, #1=5/2-2C#. Solving for #C# gives #3/4#.

Knowing #A# and #C#, we can easily solve for #B# in this equation #2x+1=5/8(x^2+4)+(Bx+3/4)(x-2)# by setting #x=1# to get #B=-5/8#.

After substituting the values in and simplifying, we have successfully decomposed the fraction to #5/(8(x-2))+(-5x+6)/(8(x^2+4))#.

One by one, we can integrate each of the distinct fractions:

#\int\ 5/(8(x-2))\ dx# #=5/8\int\ 1/u\ du# (substituting #u=x-2# and #du=dx#) #=5/8ln|u|+C# #=5/8ln|x-2|+C# (substituting #u=x-2# back)

#\int\ (-5x+6)/(8(x^2+4))\ dx# #=-5/8\int\ (x-6/5)/(x^2+4)\ dx# #=-5/8(\int\ x/(x^2+4)\ dx-\int\ (6/5)/(x^2+4)\ dx)# #=-5/8(\int\ x/(2xu)\ du-\int\ (6/5)/(x^2+4)\ dx)# (substituting #u=x^2+4# and #du=2x\ dx#) #=-5/8(1/2ln|u|+C-\int\ (6/5)/(x^2+4)\ dx)# #=-5/8(1/2ln|x^2+4|+C-\int\ (6/5)/(x^2+4)\ dx)# (substituting #u=x^2+4# back) #=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(x^2+4)\ dx)# #=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(4(tan^2(\theta)+1))\ 2sec^2(\theta)\ d\theta)# (substituting #\2tan(\theta)=x# and #dx=2sec^2(\theta)\ d\theta#) #=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(4sec^2(\theta))\ 2sec^2(\theta)\ d\theta)# #=-5/8(1/2ln|x^2+4|+C-3/5\theta)# #=-5/8(1/2ln|x^2+4|+C-3/5tan^-1(x/2))# (since #x/2=tan(\theta)#) #=3/8tan^-1(x/2)-5/16ln|x^2+4|+C#

Finally, we add the two integrals to get the final answer: #5/8ln|x-2|+3/8tan^-1(x/2)-5/16ln|x^2+4|+C#