Question

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

Answer 1

`I=x^2/2-3/2ln|x^2+3|+1/(sqrt(3))tan^-1(x/sqrt(3))+c`

Explanation:

Here,
`(x^3+1)/(x^2+3)=(x^3+3x-3x+1)/(x^2+3)=(x^3+3x)/(x^2+3)-(3x)/(x^2+3)+1/(x^2+3)`
`(x^3+1)/(x^2+3)=(x(x^2+3))/(x^2+3)-3/2*(2x)/(x^2+3)+1/(x^2+3)`
`I=int(x^3+1)/(x^2+3)dx`
`I=intxdx-3/2int(2x)/(x^2+3)dx+int1/(x^2+3)dx`
`I=x^2/2-3/2int(d/(dx)(x^2+3))/(x^2+3)dx+int1/(x^2+(sqrt(3))^2)dx`
`I=x^2/2-3/2ln|x^2+3|+1/(sqrt(3))tan^-1(x/sqrt(3))+c`