How to solve the integral: $\int {\sqrt{1+\sqrt{x}}}/x dx$ 
$$\int \frac{{\sqrt{1+\sqrt{x}}}}{x} dx$$

I tried with $u=\sqrt x $, but this did not work. I really don't know what to do...
 A: Let $u=\sqrt{1+\sqrt{x}}$ to get $x=(u^2-1)^2$ and $dx=4u(u^2-1)du$, so
$\displaystyle\int\frac{\sqrt{1+\sqrt{x}}}{x}dx=\int\frac{4u^2}{u^2-1}du=4\int\left(1+\frac{1}{u^2-1}\right)du$
A: It looks like you should be able to eliminate both problem radicals with the substitution
$$x=\tan^4t,dx=4\tan^3t\sec^2tdt$$
$$\int\dfrac{\sqrt{1+\sqrt{x}}}{x}dx=\int\frac{\sec t}{\tan^4 t}4\tan^3t\sec^2tdt=$$
$$\int\frac{4\sec^3tdt}{\tan t}=4\int\frac{dt}{\sin t\cos^2t}=4\int\frac{\sin tdt}{\cos^2t(1-\cos^2t)}$$
$$u=\cos t,du=-\sin tdt$$
$$-4\int\frac{du}{u^2(1-u^2)}=-4\int\frac{du}{u^2}-4\int\frac{du}{1-u^2}=$$
$$-4\int\frac{du}{u^2}-2\int\frac{du}{1-u}-2\int\frac{du}{1+u}$$
So far, it's been relatively straightforward, but back-substitution may be an issue.  We have
$$\frac4u+2\ln(1-u)-2\ln(1+u)=4\sec t+2\ln\frac{1-\cos t}{1+\cos t}=$$
$$4\sec t+2\ln\frac{\sec t-1}{\sec t+1}$$
From here, you'd have to use $\sec t=\sqrt{1+\sqrt x}$.  You could clean up the numerator of that natural log by rewriting it as $\ln\dfrac{\tan^4t}{(\sec t+1)^4}$, but that denominator is still going to be ugly...
