How do you find the antiderivative of int 1/(x^2+10x+21) dx?

1 Answer
George C.
Oct 12, 2016

int 1/(x^2+10x+21) dx = 1/4 ln abs(x+3) + 1/4 ln abs(x+7) + C

Explanation:

1/(x^2+10x+21) = 1/((x+3)(x+7))

color(white)(1/(x^2+10x+21)) = 1/4(((x+7)-(x+3))/((x+3)(x+7)))

color(white)(1/(x^2+10x+21)) = 1/4(1/(x+3)-1/(x+7))

So:

int 1/(x^2+10x+21) dx = int 1/4(1/(x+3)-1/(x+7)) dx

color(white)(int 1/(x^2+10x+21) dx) = 1/4 ln abs(x+3) + 1/4 ln abs(x+7) + C

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