How do you find the integral $ln(x^2 + 2x + 2)dx$ ?

Question

6174 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-15T06:30:46

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

First complete the square

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

Let

$x+1=u$ , $=>$ , $du=dx$

Therefore,

$I=intln(x^2+2x+2)dx=intln((x+2)^2+1)dx$

$=intln(u^2+1)du$

$p=ln(u^2+1)$ , $=>$ , $p'=(2u)/(u^2+1)$

$q'=1$ , $=>$ , $q=u$

Therefore,

$I=u ln(u^2+1)-int(2u^2du)/(u^2+1)$

$int(2u^2du)/(u^2+1)= 2int((u^2+1-1)du)/(u^2+1)$

$=2intdu-2int(du)/(u^2+1)$

$=2u-2arctan(u)$

Finally,

$I=u ln(u^2+1)-2u+2arctan(u)$

$=(x+1)ln(x^2+2x+2)-2(x+1)+2arctan(x+1)+C$

How do you find the integral ln(x^2 + 2x + 2)dxln(x^2 + 2x + 2)dx?