# How do you integrate int (x+5)/(2x+3) using substitution?

Aug 23, 2016

$= \frac{7}{4} \ln \left(2 x + 3\right) + \frac{1}{2} x + C$

#### Explanation:

We can't immediately substitute into this integrand. First we have to get it into a more receptive form:

We do this with polynomial long division. It's a very simple thing to do on paper but the formatting is quite difficult on here.

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

$= \frac{7}{2} \int \frac{\mathrm{dx}}{2 x + 3} + \frac{1}{2} \int \mathrm{dx}$

Now for the first integral set $u = 2 x + 3 \implies \mathrm{du} = 2 \mathrm{dx}$

$\implies \mathrm{dx} = \frac{\mathrm{du}}{2}$

$= \frac{7}{4} \int \frac{\mathrm{du}}{u} + \frac{1}{2} \int \mathrm{dx}$

$= \frac{7}{4} \ln \left(u\right) + \frac{1}{2} x + C$

$= \frac{7}{4} \ln \left(2 x + 3\right) + \frac{1}{2} x + C$