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

Answer 1

Ezekiel Alexander

#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#

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Answer 2

Isaiah Allen

To integrate ( \int \frac{2x - 5}{x^2 + 2x + 2} , dx ), perform partial fraction decomposition first, which yields ( \frac{2x - 5}{x^2 + 2x + 2} = \frac{Ax + B}{x^2 + 2x + 2} ). Solve for ( A ) and ( B ), then integrate each term separately. The result is ( \ln|x^2 + 2x + 2| - 5\sqrt{2} \tan^{-1}\left(\frac{x + 1}{\sqrt{2}}\right) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.