How do you integrate #(x^3+4)/(x^2+4)# using partial fractions?

1 Answer
Sep 2, 2016

#x^2/2-2ln(x^2+4)+2arc tan (x/2)+C#.

Explanation:

Let #I=int(x^3+4)/(x^2+4)dx#

Since the degree of poly. in Nr. #># that of in DR. , we have

to first perform long division. Instead, we proceed as under :

#Nr.=x^3+4#

#=x^3+4x-4x+4=x(x^2+4)-4x+4#. Hence,

#(x^3+4)/(x^2+4)={x(x^2+4)-4x+4}/(x^2+4)#

#=(x(x^2+4))/((x^2+4))-(4x)/(x^2+4)+4/(x^2+4)#

#=x-(4x)/(x^2+4)+4/(x^2+4)#

#:. I=int[x-(4x)/(x^2+4)+4/(x^2+4)]dx#

#=intxdx-2int(2x)/(x^2+4)dx+4int1/(x^2+4)dx#

#=x^2/2-2int(d(x^2+4))/(x^2+4)+4*1/2*arc tan (x/2)#

#=x^2/2-2ln(x^2+4)+2arc tan (x/2)+C#.

Enjoy maths.!