Question

How do you integrate `f(x)=(x^2+1)/((x^2+6)(x-4))` using partial fractions?

Answer 1

`F(x)=5/44ln(x^2+6)+10/(11sqrt6)arctan(x/sqrt6)+17/22ln|x-4|+c`

Explanation:

`f(x)=int(x^2+1)/((x^2+6)(x-4))dx`

Partial fraction decomposition is `(Ax+B)/(x^2+6)+C/(x-4)`

`(Ax+B)(x-4) + C(x^2+6) = x^2+0x+1`
From `x^2` terms: `A+C=1`
From `x^1` terms: `B-4A=0`
From `x^0` terms: `6C-4B=1`
Simultaneous solving yields `A=5/22`, `B=20/22` and `C=17/22`
Alternatively, use cover up rule where possible.

Thus the partial fraction decomposition is
`(5x+20)/(22(x^2+6))+17/(22(x-4))`

Thus `f(x)=int(5x+20)/(22(x^2+6))+17/(22(x-4))dx`

Algebraic manipulation yields:
`f(x)=5/44int(2x)/(x^2+6)dx+10/11int1/(x^2+6)dx+17/22int1/(x-4)dx`

Integrating yields:
`F(x)=5/44ln(x^2+6)+10/(11sqrt6)arctan(x/sqrt6)+17/22ln|x-4|+c`