# How do you use partial fractions to find the integral int (x+1)/(x^2+4x+3) dx?

Mar 4, 2017

$\int \frac{x + 1}{{x}^{2} + 4 x + 3} \mathrm{dx}$

We can begin by factoring the denominator of the integrand:

$\implies \int \frac{x + 1}{\left(x + 1\right) \left(x + 3\right)} \mathrm{dx}$

Cancel like terms:

$\implies \int \frac{\cancel{\left(x + 1\right)}}{\cancel{\left(x + 1\right)} \left(x + 3\right)} \mathrm{dx}$

$\implies \int \frac{1}{x + 3} \mathrm{dx}$

We can now use a simple substitution:

$u = x + 3 , \mathrm{du} = \mathrm{dx}$

$\implies \int \frac{1}{u} \mathrm{du}$

$\implies \ln \left(| u |\right) + C$

Substituting back in:

$\implies \ln \left(| x + 3 |\right) + C$