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

Answer 1

William Abdalla

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

= #9/20ln|x-2|+7/20ln|x+2|+1/10ln|x^2+1|+1/5tan^(-1)x#

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

Let #(x^3+1)/((x^2-4)(x^2+1))=A/(x-2)+B/(x+2)+(Cx+D)/(x^2+1)#

or #(x^3+1)=A(x+2)(x^2+1)+B(x-2)(x^2+1)+(Cx+D)(x+2)(x-2)#

when #x=2# we have #9=20A# i.e. #A=9/20#

when #x=-2# we have #-7=-20B# i.e. #B=7/20#

Comparing coefficients of #x^3# and constant term we get

#1=A+B+C# i.e. #C=1-A-B=1-9/20-7/20=4/20=1/5#

and #1=2A-2B-4D# i.e. #D=-(1-18/20+14/20)/4=1/5#

Hence #(x^3+1)/((x^2-4)(x^2+1))=9/(20(x-2))+7/(20(x+2))+(x+1)/(5(x^2+1))# and

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

= #int9/(20(x-2))dx+int7/(20(x+2))dx+int(x+1)/(5(x^2+1))dx#

= #9/20ln|x-2|+7/20ln|x+2|+1/5(1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx)#

= #9/20ln|x-2|+7/20ln|x+2|+1/10ln|x^2+1|+1/5tan^(-1)x#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.