Question

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

Answer 1

`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`