# 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 = $\int {2\sec^{3}\theta}d\theta$

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 = $\sin^{-1}{\sqrt{\frac{x+1}{x}}}+{\sqrt{x^2+x}}+c$

• Typo: $dx=-\dfrac{2t}{(t^2-1)^{2}}dt$. Apr 4, 2013 at 11:48

One hint may be to set $t=\sqrt{\frac{x+1}{x}}$

• @ Babak S.: great hint indeed. Thanks !! Apr 4, 2013 at 11:45
• You'r welcome :-) Apr 4, 2013 at 16:34
• Nice! ;-) $+1{}{}{}{}$ Apr 5, 2013 at 1:31

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.

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.