# Integrate\int \frac{x+3}{x \sqrt{x^{2}+6x}}dx

## Question:

Integrate {eq}\int \frac{x+3}{\sqrt{x^{2}+6x}}dx {/eq}

## Substitution Rule

We can perform a u-substitution to approximate integrals containing a function and its derivative. In this case, we assign an expression to u, and using derivatives, exchange dx for du. This derives directly from the chain rule and relates integrals to derivatives through antidifferentiation.

Here, we choose

{eq}\displaystyle u = x^2 + 6x {/eq}.

If we do this, we see that

{eq}\displaystyle \begin{align*} \frac{du}{dx} &= 2x + 6 \\ \frac{du}{2} &= (x + 3) \ dx \end{align*} {/eq}

Then we obtain

{eq}\displaystyle \begin{align*} \int \frac{x+3}{\sqrt{x^{2}+6x}}dx &= \int \frac{du}{2\sqrt{u}} \\ &= \frac{1}{2}\int u^{-1/2} \ du \\ &= u^{1/2} + C \\ &= \sqrt{x^2 + 6x} + C \end{align*} {/eq}