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

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Answer 1 · 2016-09-02T10:13:02

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

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.!