Question

Integrate :-
$\displaystyle \int \dfrac{\sqrt{x}}{1+x}\:dx$
$\: \: \: \:$
​

Answer 1

Explanation :

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

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \times \int \frac{ {t}^{2} }{1 + {t}^{2} } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \times \int \frac{ {t}^{2} + 1 - 1}{1 + {t}^{2} } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \bold{ 2 \times \int \frac{ {t}^{2} + 1 }{1 + {t}^{2} } - \frac{1}{1 + {t}^{2} } \: dt}} \\ \\ \end{gathered}\end{gathered} \end{gathered}$

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

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

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

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

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