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Factorization

∫ a+x x d x

Question

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

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Solution

$Let I = \int \sqrt{\frac{a + x}{x}} d x = \int \frac{\sqrt{(a + x) (a + x)}}{\sqrt{x (a + x)}} = \int (\frac{a + x}{\sqrt{x^{2} + a x}}) d x = a \int \frac{1}{\sqrt{x^{2} + a x}} d x + \int \frac{x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{x^{2} + a x + {(\frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \int \frac{x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int (\frac{2 x + a - a}{\sqrt{x^{2} + a x}}) d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{(2 x + a)}{\sqrt{x^{2} + a x}} d x - \frac{a}{2} \int \frac{1}{\sqrt{x^{2} + a x}} d x = \frac{a}{2} \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{(2 x + a)}{\sqrt{x^{2} + a x}} d x Putting x^{2} + a x = t in the Ist integral \Rightarrow (2 x + a) d x = d t ∴ I = \frac{a}{2} \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = \frac{a}{2} \ln |x + \frac{a}{2} + \sqrt{x^{2} + a x}| + \frac{1}{2} \times 2 \sqrt{t} + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \frac{a}{2} \ln |x + \frac{a}{2} + \sqrt{x^{2} + a x}| + \sqrt{x^{2} + a x} + C [∵ t = x^{2} + a x]$

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