# How do you find the antiderivative of int 1/(x^2+10x+25) dx?

Oct 21, 2016

$- \frac{1}{x + 5} + C$

#### Explanation:

Notice the denominator can be factored:

$\int \frac{1}{{x}^{2} + 10 x + 25} \mathrm{dx} = \int \frac{1}{x + 5} ^ 2 \mathrm{dx} = \int {\left(x + 5\right)}^{-} 2 \mathrm{dx}$

Now we can use substitution. Let $u = x + 5$. Thus, $\mathrm{du} = \mathrm{dx}$. Substituting them in:

$= \int {u}^{-} 2 \mathrm{du}$

Now use the power rule for integration $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$

$= {u}^{- 2 + 1} / \left(- 2 + 1\right) + C = {u}^{- 1} / \left(- 1\right) + C = - \frac{1}{u} + C = - \frac{1}{x + 5} + C$