Integrating $\int \sqrt{\frac{1+x}{x}}dx$ Integrating $\int\sqrt{\frac{1+x}{x}}dx$
Let, $x=\tan^{2}\theta$
$dx=2\tan\theta \sec^{2}\theta d\theta$
Integral = $\int \frac{\sec\theta}{\tan\theta}{2\tan\theta\sec^{2}\theta}d\theta$
Integral = $\int {2\sec^{3}\theta}d\theta$
 A: Let, $\sqrt\frac{x+1}{x}=t$
$\frac{x+1}{x}=t^{2}$
$1+\frac{1}{x}=t^{2}$
$\frac{1}{x}=t^{2}-1$
$\frac{1}{t^{2}-1}=x$
$dx=-\frac{2t}{(t^2-1)^{2}}dt$
Integral = -$\int\frac{2t^2}{(t^2-1)^{2}}dt$
Integral = -$\int\frac{2(t^2-1)+2}{(t^2-1)^{2}}dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{2}{(t^2-1)^{2}})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{2}{(t+1)^{2}(t-1)^{2}})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +2(\frac{1}{(t+1)(t-1)})^{2})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)}-\frac{1}{(t-1)})^{2})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)^2}+\frac{1}{(t-1)^2}-\frac{2}{(t^2-1)}))dt$
Integral = -$\int(\frac{1}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)^2}+\frac{1}{(t-1)^2}))dt$
Integral = $\sin^{-1}{t}+\frac{t}{t^2-1}+c$
Integral = $\sin^{-1}{\sqrt{\frac{x+1}{x}}}+{\sqrt{x^2+x}}+c$
A: One hint may be to set $t=\sqrt{\frac{x+1}{x}}$
A: My calculus text book says ∫(sec x)^3 dx = (1/2) sec(x) tan(x) + (1/2) ∫sec x dx
The integral of the secant function can be found here.
http://en.wikipedia.org/wiki/Integral_of_the_secant_function
I think that might be part of the solution at least. 
A: Use integration by parts to solve for $\int \sec^3 \theta d \theta$
$$u = \sec \theta$$
$$dv = \sec^2 \theta d \theta$$
$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta$$
$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta (\sec^2 \theta - 1)$$
$$\int \sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec^3 \theta d \theta + \int \sec \theta d \theta $$
$$ 2\int \sec^3 \theta d\theta = \sec \theta \tan \theta + \int \sec \theta d \theta$$
$$ \int \sec^3 \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln\left|sec \theta + \tan \theta \right| + C$$
$$ 2\int \sec^3 \theta d \theta = \sec \theta \tan \theta + \ln\left|sec \theta + \tan \theta \right| + C$$
I used parts of my answer in this question.
