Question

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

Answer 1

\[\text{ Let I }= \int\left( \frac{x + 1}{\sqrt{x^2 + 1}} \right) dx\]
` = ∫  {x      dx}/{\sqrt{x^2 + 1}} +  ∫  {dx}/{\sqrt{x^2 + 1}}`
\[\text{ Putting, x }^2 + 1 = t\]
\[ \Rightarrow \text{ 2x dx } = dt\]
\[ \Rightarrow \text{ x dx  }= \frac{dt}{2}\]
\[\text{ Then,} \]
\[I = \frac{1}{2}\int\frac{dt}{\sqrt{t}} + \int\frac{dx}{\sqrt{x^2 + 1}}\]
\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \int\frac{dx}{\sqrt{x^2 + 1}}\]
\[ = \frac{1}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \text{ log }\left| x + \sqrt{x^2 + 1} \right| + C\]
\[ = \sqrt{t} + \text{ log }\left| x + \sqrt{x^2 + 1} \right| + C\]
\[ = \sqrt{x^2 + 1} + \text{ log }\left| x + \sqrt{x^2 + 1} \right| + C\]