Question

How do you find the integral of `(6x+1)/(x^2+2x+3)`?

Answer 1

`I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c`

Explanation:

`I=int(6x+1)/(x^2+2x+3)dx`

`=int(6x+6)/(x^2+2x+3)dx-int5/(x^2+2x+3)dx`

`I=3int(2x+2)/(x^2+2x+3)dx-int5/(x^2+2x+1+2)dx`

`I=3int(d/(dx)(x^2+2x+3))/(x^2+2x+3)dx-int5/((x+1)^2+(sqrt(2))^2)dx`

`I=3ln|x^2+2x+3|-5*1/sqrt(2)tan^-1((x+1)/sqrt(2))+c`
`I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c`