how do you get the integral of dx/ x sqrt[1+ln[x]]?

Question

the dx on the top throws me off

Answer 1

use substitution:

u= 1+ln(x)
du = dx/x

so
∫ dx/ x sqrt[1+ln[x]] = ∫ du/sqrt(u) = ∫ u^{-1/2} du
= u^{1/2} / (1/2) + C = 2 u^{1/2} + C
= 2(1+ln(x) )^{1/2} + C .

Answer 2

Try not to let the dx on top throwing you off. I'll rewrite it for you, without the dx on top.

Integral (1/[x * sqrt(1 + lnx)]) dx

The reason why the dx was on top was because it was multiplied by the 1 on top. I'll rewrite this further so something can be made absolutely clear. I'm going to write it as a product.

Integral ([1/sqrt(1+lnx)][1/x]dx)

To solve this, we have to use u substitution.

Let u = 1 + lnx. Then,
du = (1/x) dx

NOW LOOK! That integral up there has just that! it has a
(1/x)dx, so we can replace all of that by du. Substitution the u, we get

Integral (1/sqrt(u)) du

Which becomes extremely clean to solve now. Note that the square root of u is equal to u^(1/2), and since it's a fraction, you really have u^(-1/2). i.e.

Integral (u^(-1/2))du

This integral is now trivial to solve.

[u^(1/2)]/[1/2] + C

Which is equal to

2u^(1/2) + C

We put back u = 1 + lnx, and we get

2(1 + lnx)^(1/2) + C

or

2 sqrt(1 + lnx) + C