What is the integral of $int (2x-5)/(x^2+2x+2)$ ?

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Answer 1 · 2018-06-29T16:36:25

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

Write $(2x-5)$ as $(2x+2-7)$

Then the integral is

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

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

$=I_1+I_2$

Calculate the integral $I_1$ by substitution

Let $u=x^2+2x+2$ , $=>$ , $du=(2x+2)dx$

$I_1=int((2x+2)dx)/(x^2+2x+2)=int(du)/u$

$=ln(u)$

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

For the integral $I_2$ , complete the square of the denominator

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

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

$I_2=int(7dx)/(x^2+2+2)=7int(dx)/((x+1)^2+1)$

$=7int(du)/(u^2+1)$

$=7arctan(u)$

$=7arctan(x+1)$

And finally,

$I=ln(x^2+2x+2)-7arctan(x+1)+C$

What is the integral of int (2x-5)/(x^2+2x+2)int (2x-5)/(x^2+2x+2)?