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

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Answer 1 · 2017-05-12T13:47:53

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

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

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

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

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

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

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

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