Calculating the integral of a logarithmic expression.

Question

The problem I have been working with is

$$\int \frac 1{\sqrt x(1+\sqrt x)}\,dx$$

The first step I did to solve this question was to set $u= 1+ \sqrt x$ the set $du = (1/2) x^{-1/2}$

Then I set $\dfrac 1 2\int\ln(du/u)$ and finally I think the answer is $\ln\dfrac 1 x+c$

Answer 1

Setting $u=1+\sqrt x$, we have $$\frac{du}{dx}=\frac{1}{2\sqrt x}\Rightarrow \frac{dx}{\sqrt x}=2du,$$ so $$\int\frac{1}{1+\sqrt x}\cdot \frac{dx}{\sqrt x}=\int \frac 2u du=2\ln u+C=2\ln(1+\sqrt x)+C.$$

Answer 2

\begin{align} u & = \sqrt x \\[10pt] u^2 & = x \\[10pt] 2u\,du & = dx \\[10pt] \int\frac{dx}{\sqrt x(1+\sqrt x)} & = \int \frac{2u\,du}{u(1+u)} = 2\int\frac{du}{1+u}\quad\text{etc.} \end{align}