# How do you integrate dx / (2sqrt(x) + 2x?

Mar 30, 2015

$\int \frac{\mathrm{dx}}{2 \sqrt{x} + 2 x}$

There are "obvious" common factors in the denominator.

What do I know about $\sqrt{x}$ ? Among other things, I know the derivative is $\frac{1}{2 \sqrt{x}}$

I could factor out $2 \sqrt{x}$ in the denominator. Will that help?
Try it and see. (Yes, I can try it in my head. Students will need to try it on paper.)

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

Did that help?

Not sure? write it as a product of 2 fracions where one fraction is the derivative of $\sqrt{x}$

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

Now, what if I let $u = 1 + \sqrt{x}$?

Then $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ and our integral becomes $\int \frac{1}{u} \mathrm{du}$ which is $\ln u$, so we can quickly finish:

$\int \frac{\mathrm{dx}}{2 \sqrt{x} + 2 x} = \ln \left(1 + \sqrt{x}\right) + C$