How do you find #int (2x-5)/(x^2+2x+2)#?

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Answer 1 · 2016-11-21T08:00:01

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

Let's rewrite

#(2x-5)/(x^2+2x+2)=(2x+2-7)/(x^2+2x+2)#

#=(2x+2)/(x^2+2x+2)-7/(x^2+2x+2)#

Therefore, the integral is

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

So, we have 2 integrals

Let's do the first one,

Let #u=x^2+2x+2#

Then #du=(2x+2)dx#

#int((2x+2)dx)/(x^2+2x+2)=int(du)/u=lnu#

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

For the second integral,

#x^2+2x+2=x^2+2x+1+1#

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

Let #u=x+1#

#du=dx#

#int(7dx)/(x^2+2x+2)=7int(du)/(u^2+1)#

#=7arctanu#

#=7arctan(x+1)#

Finally, we have

#int((2x-5)dx)/(x^2+2x+2)=ln(x^2+2x+2)-7arctan(x+1)+C#