How do you find #int (x^2-4)/((x+3)(x^2+1))dx# using partial fractions?

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Answer 1 · 2018-05-06T05:00:54

# 1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C#.

Suppose that, #I=int(x^2-4)/{(x+3)(x^2+1)}dx#.

Let, #(x^2-4)/{(x+3)(x^2+1)}=A/(x+3)+(Bx+C)/(x^2+1), (A,B,C in RR)...(ast)#.

We will use Heavyside's Method to determine #A,B,C#.

#:. A=[(x^2-4)/(x^2+1)]_(x=-3)=(9-4)/(9+1)=1/2#.

Using #A# in #(ast)#, we have,

# (Bx+C)/(x^2+1)=(x^2-4)/{(x+3)(x^2+1)}-(1/2)/(x+3)#,

#={2(x^2-4)-(x^2+1)}/{2(x+3)(x^2+1)}, i.e., #

# (Bx+C)/(x^2+1)=(x^2-9)/{2(x+3)(x^2+1)}={1/2(x-3)}/(x^2+1)#.

#:. B=1/2, and, C=-3/2#.

Altogether, we have,

#(x^2-4)/{(x+3)(x^2+1)}=(1/2)/(x+3)+(1/2x-3/2)/(x^2+1)#.

#:. I=1/2int1/(x+3)dx+1/2intx/(x^2+1)dx-3/2int1/(x^2+1)dx#,

#=1/2ln|(x+3)|+1/4int(2x)/(x^2+1)dx-3/2arc tanx#,

#=1/2ln|(x+3)|+1/4int{d/dx(x^2+1)}/(x^2+1)dx-3/2arc tanx#,

#rArr I=1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C#.

Enjoy Maths.!