Question

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

Answer 1

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

Explanation:

Let's perform a long division

`color(white)(aaaa)` `x^4` `color(white)(aaaaaa)` `x-4` `color(white)(aaaa)` `|` `x^2+2`

`color(white)(aaaa)` `x^4+2x^2` `color(white)(aaaa)` ###color(white)(aaaaa)#`|` `x^2-2`

`color(white)(aaaaa)` `0+2x^2+x` `color(white)(aa)` `-4`

`color(white)(aaaaaaaa)` `+0+x` `color(white)(aaaa)` `0`

Therefore,

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

So,

`int((x^4+x-4)dx)/(x^2+2)=intx^2dx-int2dx+int(xdx)/(x^2+2)`

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

For the last integral,

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

`int(xdx)/(x^2+2)=1/2int(du)/u`

`=1/2lnu`

`=1/2ln(x^2+2)`

Putting it all together

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