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

1 Answer
Jun 29, 2018

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

Explanation:

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#