# How do you use substitution to integrate (2x+7)/(x^2+5x+6) dx?

Jun 16, 2015

The answer can be written as $\int \frac{2 x + 7}{{x}^{2} + 5 x + 6} \setminus \mathrm{dx} = \ln | \frac{{\left(x + 2\right)}^{3}}{x + 3} | + C$

#### Explanation:

You don't need substitution to do this integral. Use the Method of Partial Fractions instead.

Set $\frac{2 x + 7}{{x}^{2} + 5 x + 6} = \frac{A}{x + 2} + \frac{B}{x + 3}$ and solve for $A$ and $B$ (note that ${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$).

Multiply both sides of the preceding equation by $\left(x + 2\right) \left(x + 3\right)$ to get, after cancellation, $2 x + 7 = A \left(x + 3\right) + B \left(x + 2\right)$. The quickest way to solve for $A$ and $B$ is to substitute $x = - 3$ and $x = - 2$ into this last equation (even though they make the original equation undefined).

$x = - 3 \setminus R i g h t a r r o w 1 = - B \setminus R i g h t a r r o w B = - 1$

$x = - 2 \setminus R i g h t a r r o w 3 = A$

Hence, $\frac{2 x + 7}{{x}^{2} + 5 x + 6} = \frac{3}{x + 2} - \frac{1}{x + 3}$

Now substitute, integrate, and use properties of logarithms:

$\int \frac{2 x + 7}{{x}^{2} + 5 x + 6} \setminus \mathrm{dx} = \int \left(\frac{3}{x + 2} - \frac{1}{x + 3}\right) \setminus \mathrm{dx}$

$= 3 \ln | x + 2 | - \ln | x + 3 | + C = \ln | \frac{{\left(x + 2\right)}^{3}}{x + 3} | + C$