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

Question

William Abdalla · Accepted Answer

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

#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#

Adam Adkins · Answer

undefined To find the antiderivative of $ \int \frac{1}{x^2 + 10x + 21} \, dx $, we first need to factor the denominator. The expression $ x^2 + 10x + 21 $ factors into $ (x + 7)(x + 3) $.

Therefore, the integral becomes $ \int \frac{1}{(x + 7)(x + 3)} \, dx $.

Next, we use partial fraction decomposition to break down the fraction into simpler fractions. We express $ \frac{1}{(x + 7)(x + 3)} $ as $ \frac{A}{x + 7} + \frac{B}{x + 3} $.

After finding the values of $ A $ and $ B $, we integrate each term separately.

Once integrated, the antiderivative of $ \frac{1}{x^2 + 10x + 21} $ will be expressed as a sum of the antiderivatives of the simpler fractions, each corresponding to the terms $ \frac{A}{x + 7} $ and $ \frac{B}{x + 3} $, plus a constant of integration.