Question

Integration!
\[\int \frac{4}{x + \sqrt{x}} dx\]

Answer 1

I can pull the 4 out, of course.
\[4\int \frac{1}{x + \sqrt{x}} dx\]
There is probably something clever I am overlooking involving rewriting everything in terms of \(\sqrt{x}\) so it becomes an expression like \(a^2 +a\)... but, I'm not sure what to do with it.

Answer 2

take the constant out and use the integral of "natural log"

Answer 3

What do I use as a u?

Answer 4

use substitution u=sqrt(x)

Answer 5

\[u = \sqrt{x}\]\[du = \frac{1}{2\sqrt{x}}\]
Since the \(\sqrt{x}\) in the denominator is added, not multiplied, I'm not seeing how to get that du to work. I'm probably missing something simple.

Answer 6

Oh, with \(\frac{1}{\sqrt{x}(\sqrt{x} + 1)}\) that might be able to work.

Answer 7

So that gives
\[4\int 2 \frac{1}{u + 1} du\]
If I want to be really proper I need a v but du = dv so I can just shove it in.
\[8 \ln |u + 1| + C\]Which gives me \[8 \ln |\sqrt{x} + 1| + C\]

Answer 8

Thanks for the hints!

Answer 9

you are welcome