Question

Integrate :-


`\displaystyle \int (√(x))/(1+x)\:dx`

`\: \: \: \:`
​

Answer 1

Explanation :

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ \int \frac{2 \: {t}^(2) }{1 + {t}^(2) } \: dt }} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 * \int \frac{ {t}^(2) }{1 + {t}^(2) } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 * \int \frac{ {t}^(2) + 1 - 1}{1 + {t}^(2) } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 * \int \frac{ {t}^(2) + 1 }{1 + {t}^(2) } - \frac{1}{1 + {t}^(2) } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \bigg( \int \frac{ {t}^(2) + 1 }{1 + {t}^(2) } \: dt - \int\frac{1}{1 + {t}^(2) } \: dt} \bigg)} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \bigg( t - \int \frac{ 1 }{1 + {t}^(2) } \: dt \bigg)}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \bigg(t - arctan(t) \bigg)}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \bigg( √(x) - arctan( √(x) ) \bigg)}} \\ \\ \end{gathered}\end{gathered} \end{gathered}`

`\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 √(x) - 2 \: arctan( √(x) \: ) }} \\ \\ \end{gathered}\end{gathered} \end{gathered}\\ \\ \begin{gathered}\begin{gathered}\begin{gathered}\\ :\implies\large{ \boxed{ \bold{ 2√(x) - 2 \: arctan( √(x) \: ) + C}}} \\\end{gathered} \\ \\\end{gathered}\end{gathered}`