Question

How do you find the indefinite integral of `int (x^3-3x^2+4x-9)/(x^2+3)`?

Answer 1

The answer is `=x^2/2-3x+1/2ln(x^2+3)+C`

Explanation:

Let's perform the long division

`color(white)(aaaa)` `x^3-3x^2+4x-9` `color(white)(aaaa)` `∣` `x^2+3`

`color(white)(aaaa)` `x^3` `color(white)(aaaaaa)` `+3x` `color(white)(aaaaaaaa)` `∣` `x-3`

`color(white)(aaaaa)` `0-3x^2+x-9`

`color(white)(aaaaaaa)` `-3x^2+x-9`

`color(white)(aaaaaaaaa)` `-0+x+0`

So,

`(x^3-3x^2+4x-9)/(x^2+3)=(x-3)+x/(x^2+3)`

`int((x^3-3x^2+4x-9)dx)/(x^2+3)=int(x-3)dx+int(xdx)/(x^2+3)`

`=x^2/2-3x+1/2int(2xdx)/(x^2+3)`

Let `u=x^2+3`, `du=2xdx`

`int(2xdx)/(x^2+3)=int(du)/u=lnu= ln(x^2+3)`

`int((x^3-3x^2+4x-9)dx)/(x^2+3)=x^2/2-3x+1/2ln(x^2+3)+C`