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

Question

1939 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T17:53:23

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

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

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

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

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

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

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