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

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Answer 1 · 2016-09-08T14:35:37

#1/6{2arc tanx-arc tan (x/2)}+C#.

Let #I=int1/((x^2+1)(x^2+4))dx#.

We will use the Method of Partial Fraction to decompose the

integrand #1/((x^2+1)(x^2+4))=1/((y+1)(y+4))#, where, #y=x^2#.

For this, we have to find #A,B in RR# such that,

#1/((y+1)(y+4))=A/(y+1)+B/(y+4)#.

Using Heavyside's Cover-up Method, we have,

#A=[1/(y+4)]_(y=-1)=1/3#.

#B=[1/(y+1)]_(y=-4)=-1/3#.

#:. 1/((y+1)(y+4))=(1/3)/(y+1)+(-1/3)/(y+4)#. Since, #y=x^2#,

#1/((x^2+1)(x^2+4))=1/3(1/(x^2+1))-1/3(1/(x^2+4))#. Therefore,

#I=1/3int(1/(x^2+1))dx-1/3int(1/(x^2+2^2))dx#

#=1/3arc tan x-1/3*1/2arc tan (x/2)#, i.e.,

#I=1/6{2arc tanx-arc tan (x/2)}+C#.

Enjoy Maths.!