<s>[INST] 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 = $\int {2\sec^{3}\theta}d\theta$
 [/INST] 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.
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