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How do you find the integral of #(6x+1)/(x^2+2x+3)#?

Answer 1

Ezra Adams

#I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c#

#I=int(6x+1)/(x^2+2x+3)dx#

#=int(6x+6)/(x^2+2x+3)dx-int5/(x^2+2x+3)dx#

#I=3int(2x+2)/(x^2+2x+3)dx-int5/(x^2+2x+1+2)dx#

#I=3int(d/(dx)(x^2+2x+3))/(x^2+2x+3)dx-int5/((x+1)^2+(sqrt(2))^2)dx#

#I=3ln|x^2+2x+3|-5*1/sqrt(2)tan^-1((x+1)/sqrt(2))+c# #I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c#

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Answer 2

Emily Ammerman

To find the integral of (\frac{{6x+1}}{{x^2+2x+3}}), you can use partial fraction decomposition. First, complete the square in the denominator to rewrite it as (x^2+2x+3 = (x+1)^2 + 2). Then, rewrite the integral using partial fractions, solve for the constants, and integrate each term separately. The result will be:

[ \int \frac{{6x+1}}{{x^2+2x+3}} dx = \int \left( \frac{Ax + B}{{(x+1)^2 + 2}} + \frac{Cx + D}{{(x+1)^2 + 2}} \right) dx ]

where (A), (B), (C), and (D) are constants to be determined. After finding the constants, integrate each term separately to get the final result.

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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.