Question

\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{\left( x + 2 \right) dx}{\sqrt{x^2 + 2x - 1}}\]
\[\text{ Consider }, \]
\[x + 2 = A \frac{d}{dx} \left( x^2 + 2x - 1 \right) + B\]
\[ \Rightarrow x + 2 = A \left( 2x + 2 \right) + B\]
\[ \Rightarrow x + 2 = \left( 2A \right) x + 2A + B\]
\[\text{ Equating Coefficients of  like terms}\]
\[2A = 1\]
\[ \Rightarrow A = \frac{1}{2}\]
\[\text{ And }\]
\[2A + B = 2\]
\[ \Rightarrow 2 \times \frac{1}{2} + B = 2\]
\[ \Rightarrow B = 1\]
\[\text{ Then }, \]
\[I = \int\frac{\left[ \frac{1}{2} \left( 2x + 2 \right) + 1 \right]}{\sqrt{x^2 + 2x - 1}}dx\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 2 \right) dx}{\sqrt{x^2 + 2x - 1}} + \int\frac{dx}{\sqrt{x^2 + 2x - 1}}\]
\[\text{ let x }^2 + 2x - 1 = t\]
\[ \Rightarrow \left( 2x + 2 \right) dx = dt\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{\sqrt{t}} + \int\frac{dx}{\sqrt{x^2 + 2x - 1}}\]
\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \int\frac{dx}{\sqrt{x^2 + 2x + 1 - 2}}\]
\[ = \frac{1}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \int\frac{dx}{\sqrt{\left( x + 1 \right)^2 - \left( \sqrt{2} \right)^2}}\]
\[ = \sqrt{t} + \text{ log  }\left| x + 1 + \sqrt{\left( x + 1 \right)^2 - \left( \sqrt{2} \right)^2} \right| + C\]
\[ = \sqrt{x^2 + 2x - 1} + \text{ log }\left| x + 1 + \sqrt{x^2 + 2x - 1} \right| + C\]