# How do you integrate int 2/sqrt(3x-7) using substitution?

Feb 17, 2017

$\left(\frac{1}{3}\right) \left(\sqrt{3 x - 7}\right)$

#### Explanation:

Set $u = 3 x - 7$

That means that $\mathrm{du} = 3 \mathrm{dx}$. However, you don't have a 3 on the top.

Don't worry. Instead, just multiply the integral by $\frac{3}{3}$. That way, you have the 3 sufficient for the substitution.

$\int \frac{3 \cdot 2}{3 \sqrt{3 x - 7}} \mathrm{dx}$

You can then substitute u into the square root and du for 3dx using the three on the top. You should then get

$\int \frac{2}{3 \sqrt{u}}$

Now personally, I like to take the $\frac{2}{3}$ out from the integral and write $\sqrt{u}$ as ${u}^{- \frac{1}{2}}$. Rewriting this, you get

$\frac{2}{3} \int {u}^{- \frac{1}{2}}$

Ignore the $\frac{2}{3}$ for now and focus on the $\int {u}^{- \frac{1}{2}}$. Add 1 to the power and then multiply by $\frac{1}{2}$ since that's the number you get from adding 1 to the power. Simplify and you should have:

$\left(\frac{1}{3}\right) \left({u}^{\frac{1}{2}}\right)$.

Now since $u = 3 x - 7$, we need to plug that back into the result. Also, the power of $\frac{1}{2}$ is the same thing as the square root. Plugging in and writing a square root, we should get:

$\left(\frac{1}{3}\right) \left(\sqrt{3 x - 7}\right)$