Let #160;I= #8747;1+xxdx #160; #160; #160; #160; #160; #160; #160; #160;= #8747;1+xx #215;1+x1+xdx #160; #160; #160; #160; #16 ...

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Factorization

∫ 1+x x d x

Question

$\int \sqrt{\frac{1 + x}{x}} d x$

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Solution

$Let I = \int \sqrt{\frac{1 + x}{x}} d x = \int \frac{\sqrt{1 + x}}{\sqrt{x}} \times \frac{\sqrt{1 + x}}{\sqrt{1 + x}} d x = \int (\frac{1 + x}{\sqrt{x^{2} + x}}) d x Let x + 1 = A \frac{d}{d x} (x^{2} + x) + B \Rightarrow x + 1 = A (2 x + 1) + B \Rightarrow x + 1 = (2 A) x + A + B Equating the coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} & A + B = 1 \Rightarrow \frac{1}{2} + B = 1 ∴ B = \frac{1}{2} ∴ I = \int \frac{(x + 1)}{\sqrt{x^{2} + x}} d x = \int (\frac{\frac{1}{2} (2 x + 1) + \frac{1}{2}}{\sqrt{x^{2} + x}}) d x = \frac{1}{2} \int \frac{(2 x + 1)}{\sqrt{x^{2} + x}} d x + \frac{1}{2} \int \frac{1}{\sqrt{x^{2} + x}} d x$

$Putting x^{2} + x = t \Rightarrow (2 x + 1) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{\sqrt{t}} d t + \frac{1}{2} \int \frac{1}{\sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \int \frac{1}{\sqrt{t}} d t + \frac{1}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \frac{1}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \ln |x + \frac{1}{2} + \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4}}| + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \sqrt{t} + \frac{1}{2} \ln |x + \frac{1}{2} + \sqrt{x^{2} + x}| + C = \sqrt{x^{2} + x} + \frac{1}{2} \ln |(x + \frac{1}{2}) + \sqrt{x^{2} + x}| + C [∵ t = x^{2} + x]$

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