# Question


$$\int \sqrt{\frac{\mathrm{a}+\mathrm{c}x}{x}} \mathrm{d}x$$

# Solution

$Let \, x = \frac{a}{c} \cdot \tan^2{\theta}$

$\Rightarrow \mathrm{d}x= \frac{2a}{c} \cdot \tan\theta \cdot \sec^2\theta \, \mathrm{d}\theta$

$a+cx = a+\not{c} \cdot \frac{a}{\not c} \tan^2{\theta} = a+a\tan^2\theta = a(1+\tan^2\theta) = a \cdot \sec^2\theta$

$\int \sqrt{\frac{\mathrm{a}+\mathrm{c}x}{x}} \mathrm{d}x = \int \sqrt{\frac{a \cdot \sec^2\theta}{\frac{a}{c} \cdot \tan^2{\theta}}}\cdot \frac{2a}{c} \cdot \tan\theta \cdot \sec^2\theta \, \mathrm{d}\theta =2 \cdot a \cdot c^{-\frac{1}{2}} \cdot \int \sec^3\theta \, \mathrm{d}\theta$

$\int \sec^3\theta \, \mathrm{d}\theta = \frac{1}{2}[\sec \theta \tan \theta + \ln \vert \tan \theta + \sec \theta \vert]$

$\Rightarrow \int \sqrt{\frac{\mathrm{a}+\mathrm{c}x}{x}} \mathrm{d}x = 2 \cdot a \cdot c^{-\frac{1}{2}} \cdot\frac{1}{2}[\sec \theta \tan \theta + \ln \vert \tan \theta + \sec \theta \vert ]$

$\tan \theta = \sqrt{\frac{cx}{a}},\, \sec \theta = \sqrt{\frac{a+cx}{a}}$

$\Rightarrow 2 \cdot a \cdot c^{-\frac{1}{2}} \cdot\frac{1}{2}[\sec \theta \tan \theta + \ln \vert \tan \theta + \sec \theta \vert ] = \not 2 \cdot a \cdot c^{-\frac{1}{2}} \cdot \frac{1}{\not 2}[\sqrt{\frac{a+cx}{a}} \cdot \sqrt{\frac{cx}{a}} + \ln \vert \sqrt{\frac{cx}{a}} + \sqrt{\frac{a+cx}{a}} \vert ]$

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$$\mathrm{Ans: } \, a \cdot c^{-\frac{1}{2}} \cdot[\sqrt{\frac{a+cx}{a}} \cdot \sqrt{\frac{cx}{a}} + \ln \vert \sqrt{\frac{cx}{a}} + \sqrt{\frac{a+cx}{a}} \vert ] + C$$