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

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

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Answer 1 · 2016-08-28T18:49:08

$ln|x|-2/x-arc tanx+C$

Let $I=int(x^3+x^2+x+2)/(x^4+x^2)dx=int(x^3+x^2+x+2)/(x^2(x^2+1))dx$

On the first hand inspection, we find that we have no go but to use

the Method of Partial Fraction to decompose the Integrand, but,

$"The Nr."$

$=(x^3+x^2+x+2)=ul(x^3+x)+ul(x^2+1)+1$

$=x(x^2+1)+1(x^2+1)+1=(x^2+1)(x+1)+1$ .

Hence, $(x^3+x^2+x+2)/(x^2(x^2+1))={(x^2+1)(x+1)+1}/(x^2(x^2+1))$

$={(x^2+1)(x+1)}/ (x^2(x^2+1))+1/(x^2(x^2+1))$

$=((x^2+1)(x+1))/(x^2(x^2+1))+1/(x^2(x^2+1))$

$=(x+1)/x^2+{(x^2+1)-x^2)/(x^2(x^2+1)$

$=x/x^2+1/x^2+(x^2+1)/(x^2(x^2+1))-x^2/(x^2(x^2+1)$

$=1/x+1/x^2+1/x^2-1/(x^2+1)$ .

Therefore,

$I=int[1/x+2/x^2-1/(x^2+1)]dx$

$=ln|x|-2/x-arc tanx+C$

Enjoy maths.!