Question

\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\sqrt{\frac{1 + x}{x}}dx\]

\[ = \int\frac{\sqrt{1 + x}}{\sqrt{x}} \times \frac{\sqrt{1 + x}}{\sqrt{1 + x}}dx\]

\[ = \int\left( \frac{1 + x}{\sqrt{x^2 + x}} \right)dx\]

\[\text{ Let  x }+ 1 = A\frac{d}{dx}\left( x^2 + x \right) + B\]

\[ \Rightarrow x + 1 = A \left( 2x + 1 \right) + B\]

\[ \Rightarrow x + 1 = \left( 2A \right)x + A + B\]

\[\text{Equating the coefficients of like terms}\]

\[2A = 1\]

\[ \Rightarrow A = \frac{1}{2}\]

\[\text{ and  A + B = 1 }\]

\[ \Rightarrow \frac{1}{2} + B = 1\]

\[ \therefore B = \frac{1}{2}\]

\[ \therefore I = \int\frac{\left( x + 1 \right)}{\sqrt{x^2 + x}}dx\]

\[ = \int\left( \frac{\frac{1}{2} \left( 2x + 1 \right) + \frac{1}{2}}{\sqrt{x^2 + x}} \right)dx\]

\[ = \frac{1}{2}\int\frac{\left( 2x + 1 \right)}{\sqrt{x^2 + x}}dx + \frac{1}{2}\int\frac{1}{\sqrt{x^2 + x}}dx\]

\[\text{ Putting x}^2 + x = t\]

\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]

\[ \therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\]

\[ = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\]

\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\]

\[ = \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \text{ ln }\left| x + \frac{1}{2} + \sqrt{\left( x + \frac{1}{2} \right)^2 - \frac{1}{4}} \right| + C............ \left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]

\[ = \sqrt{t} + \frac{1}{2} \text{ ln} \left| x + \frac{1}{2} + \sqrt{x^2 + x} \right| + C\]

\[ = \sqrt{x^2 + x} + \frac{1}{2} \text{ ln} \left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x} \right| + C................... \left[ \because t = x^2 + x \right]\]