What is the integration of \[ \int {\frac{x}{{\sqrt x + 1}}} dx\] ~~~~~[NCERT Exemp. Q. 10,Page 164 ]

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

Let’s put \[\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}dx = dt\]

\[ \Rightarrow \] \[dx = 2\sqrt x dt\]

therefore,\[I = 2\int {\left( {\frac{{x\sqrt x }}{{t + 1}}} \right)} dt = 2\int {\frac{{{t^2} \cdot t}}{{t + 1}}} dt = 2\int {\frac{{{t^3}}}{{t + 1}}} dt\]


\[ = 2\int {\frac{{{t^3} + 1 – 1}}{{t + 1}}} dt = 2\int {\frac{{(t + 1)\left( {{t^2} – t + 1} \right)}}{{t + 1}}} dt – 2\int {\frac{1}{{t + 1}}} dt\]

\[ = 2\int {\left( {{t^2} – t + 1} \right)} dt – 2\int {\frac{1}{{t + 1}}} dt\]


\[ = 2\left[ {\frac{{{t^3}}}{3} – \frac{{{t^2}}}{2} + t – \log |(t + 1)|} \right] + C\]


\[ = 2\left[ {\frac{{x\sqrt x }}{3} – \frac{x}{2} + \sqrt x – \log |(\sqrt x + 1)|} \right] + C\]

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