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

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Answer 1 · 2018-03-24T17:55:38

#I=ln|x^2+2x+2|-7tan^-1(x+1)+c#

We know that

#color(red)((1) int(f^'(x))/(f(x))dx=ln|f(x)|+c#

#color(red)((2)int1/(U^2+1)dx=tan^-1U+c#

Now

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

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

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

#=int(d/(dx)(x^2+2x+2))/(x^2+2x+2)dx-7int1/(x^2+2x+1+1)dx#

#I=int(d/(dx)(x^2+2x+2))/(x^2+2x+2)dx-7int1/((x+1)^2+1)dx#

Using #color(red)((1) and (2),# we get

#I=ln|x^2+2x+2|-7tan^-1(x+1)+c#