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

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Answer 1 · 2017-01-08T10:58:07

The answer is #=x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C#

Since the degree of the numerator is not less than the degree of the denominator, perform a long division

#color(white)(aaaa)##x^2+3x##color(white)(aaaaaaaa)##∣##x^2-4#

#color(white)(aaaa)##x^2##color(white)(aaaaaa)##-4##color(white)(aaaa)##∣##1#

#color(white)(aaaa)##0+3x##color(white)(aaa)##+4#

Therefore,

By factorising the denominator,

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

Now, we perform the partial fraction decomposition

#(3x+4)/((x+2)(x-2))=A/(x+2)+B/(x-2)#

#=(A(x-2)+B(x+2))/((x+2)(x-2))#

So,

#3x+4=A(x-2)+B(x+2)#

Let #x=2#, #=>#, #10=4B#, #=>#, #B=5/2#

Let #x=-2#, #=>#, #-2=-4A#, #=>#, #A=1/2#

Therefore,

#(x^2+3x)/(x^2-4)=1+(1/2)/(x+2)+(5/2)/(x-2)#

So,

#int((x^2+3x)dx)/(x^2-4)=int1dx+(1/2)intdx/(x+2)+(5/2)intdx/(x-2)#

#=x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C#