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

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Answer 1 · 2018-03-19T14:48:21

$1/2ln|x^2+1|+1/sqrt2tan^-1(x/sqrt2)+c$

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

Rearranging the terms in numerator,

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

Simplifying according to required factors in denominator,

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

$=int(xcancel((x^2+2)))/((x^2+1)cancel((x^2+2)))dx+intcancel((x^2+1))/(cancel((x^2+1))(x^2+2))dx$

$=intx/(x^2+1)dx+int1/(x^2+2)dx$

$=1/2int(2x)/(x^2+1)dx+int1/(x^2+(sqrt2)^2)dx$

$=1/2int(d/(dx)(x^2+1))/(x^2+1)dx+int1/(x^2+(sqrt2)^2)dx$

$=1/2ln|x^2+1|+1/sqrt2tan^-1(x/sqrt2)+c$

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