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

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

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