Question

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

Answer 1

`int dx / (2sqrt(x) + 2x )`

Think about integration techniques, think about derivatives, think about fractions and factoring and . . .

There are "obvious" common factors in the denominator.

What do I know about `sqrtx` ? Among other things, I know the derivative is `1/(2sqrtx)`

I could factor out `2sqrtx` 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 dx / (2sqrt(x) + 2x ) = int 1/(2sqrtx(1+sqrtx))dx`

Did that help?

Not sure? write it as a product of 2 fracions where one fraction is the derivative of `sqrtx`

`int dx / (2sqrt(x) + 2x ) = int 1/(2sqrtx) 1/(1+sqrtx)dx`

Now, what if I let `u=1+sqrtx`?

Then `du = 1/(2sqrtx) dx` and our integral becomes `int 1/u du` which is `lnu`, so we can quickly finish:

`int dx / (2sqrt(x) + 2x ) = ln(1+sqrtx) +C`