Answer
$$\int \frac{\sqrt{1+\sqrt{x}}}{x} d x=4 \sqrt{1+\sqrt{x}}+2 \ln \frac{| \sqrt{1+\sqrt{x}}-1|}{|1+\sqrt{1+\sqrt{x}}|}+C$$
Work Step by Step
Given $$\int \frac{\sqrt{1+\sqrt{x}}}{x} d x$$
Let
\begin{aligned} u &=\sqrt{1+\sqrt{x}} \\ u^{2} &=1+\sqrt{x} \Rightarrow x=(u^2-1)^2\\ d x &=2\left(u^{2}-1\right)(2 u) \\
d u&=4 u\left(u^{2}-1\right) d u \end{aligned}
So, we get
\begin{aligned}I&=
\int \frac{\sqrt{1+\sqrt{x}}}{x} d x \\
&=\int \frac{u}{\left(u^{2}-1\right)^{2}} 4 u\left(u^{2}-1)\right.du
\\ &=4 \int \frac{u^{2}}{u^{2}-1} d u\\
& =4 \int \frac{u^{2}-1+1}{u^{2}-1} d u\\
&=4 \int \left(\frac{u^{2}-1}{u^{2}-1}+\frac{1}{u^{2}-1}\right) d u \\
&=4 \int\left(1+\frac{1}{u^{2}-1}\right) d u \end{aligned}
Since
\begin{aligned} \frac{1}{u^{2}-1} &=\frac{A}{u-1}+\frac{B}{u+1} \\ 1 &=A(u+1)+B(u-1) \\ \text { at } u=-1 : \quad & 1=0+B(-1-1)=-2 B \\ B &=-\frac{1}{2} \\ \text { at } u=1 : \quad & 1=A(1+1)+0=2 A \\ A &=\frac{1}{2} \end{aligned}
So, we get
\begin{aligned} I&=4 \int\left(1+\frac{1}{u^{2}-1}\right) d u \\
&=4 \int\left(1+\frac{1 / 2}{u-1}+\frac{-1 / 2}{u+1}\right) d u \\ &=4\left(u+\frac{1}{2} \ln |u-1|-\frac{1}{2} \ln |u+1|\right)+C \\ &
=4 u+2 \ln |u-1|-2 \ln |u+1|+C \\
&=4\left(u+\frac{1}{2} \ln\frac{ |u-1|}{ |u+1|}\right)+C \\ &
=4 u+2 \ln |u-1|-2 \ln |u+1|+C \\
&=4 \sqrt{1+\sqrt{x}}+2 \ln \frac{| \sqrt{1+\sqrt{x}}-1|}{|1+\sqrt{1+\sqrt{x}}|}+C\\\end{aligned}
