Question

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

Answer 1

`26/11ln|x-3|+7/22ln(x^2+2)+21/(11sqrt2)tan^(-1)(x/sqrt2)+K`.

Explanation:

We split the Integrand as :
`(3x^2-1)/((x^2+2)(x-3))=A/(x-3)+(Bx+C)/(x^2+2),......................(star)`
where, `A,B,C in RR`.

`A` can be quickly determined by Heavyside"s Cover-up Method as

`A=[(3x^2-1)/(x^2+2)]_(x=3)=26/11`

Simplifying `(star)` & comparing rational polys. on both sides, we get,
`A(x^2+2)+(Bx+C)(x-3)=3x^2-1.....................(1)`

Taking `x=0, &, A=26/11` in `(1)`,
`26/11(2)+C(-3)=-1rArrC=21/11`.

`(1), x=1, A=26/11, C=21/11`
`rArr26/11(3)+(B+21/11)(-2)=2rArrB=7/11`.

With these `A,B,C`, we have, `int(3x^2-1)/((x^2+2)(x-3))dx`,

`=26/11int1/(x-3)dx+1/11int(7x+21)/(x^2+2)dx`,

`=26/11ln|x-3|+7/11*1/2int(2x)/(x^2+2)dx+21/11int1/(x^2+2)dx`,

`=26/11ln|x-3|+7/22int(d(x^2+2))/(x^2+2)+21/11*1/sqrt2*tan^(-1)(x/sqrt2)`,

`=26/11ln|x-3|+7/22ln(x^2+2)+21/(11sqrt2)tan^(-1)(x/sqrt2)+K`.

Enjoy Maths!