Question

How do you find `int (2x-5)/(x^2+2x+2)`?

Answer 1

The answer is `=ln(x^2+2x+2)-7arctan(x+1)+C`

Explanation:

Let's rewrite

`(2x-5)/(x^2+2x+2)=(2x+2-7)/(x^2+2x+2)`

`=(2x+2)/(x^2+2x+2)-7/(x^2+2x+2)`

Therefore, the integral is

`int((2x-5)dx)/(x^2+2x+2)=int((2x+2)dx)/(x^2+2x+2)-int(7dx)/(x^2+2x+2)`

So, we have 2 integrals

Let's do the first one,

Let `u=x^2+2x+2`

Then `du=(2x+2)dx`

`int((2x+2)dx)/(x^2+2x+2)=int(du)/u=lnu`

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

For the second integral,

`x^2+2x+2=x^2+2x+1+1`

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

Let `u=x+1`

`du=dx`

`int(7dx)/(x^2+2x+2)=7int(du)/(u^2+1)`

`=7arctanu`

`=7arctan(x+1)`

Finally, we have

`int((2x-5)dx)/(x^2+2x+2)=ln(x^2+2x+2)-7arctan(x+1)+C`