Question

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

Answer 1

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

Explanation:

`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`