Question

Integrate :-$\displaystyle \bf \large\int\limits^1_0\dfrac{log (x+)}{x} dx$​

Answer 1

Answer:

GIVEN :-

$\begin{gathered}\begin{gathered}\\ \bullet\:{ \bold{ \int \dfrac{\sqrt{x}}{1+x}\:dx}}\\ \\ \end{gathered}\end{gathered}$

TO FIND :-

$\begin{gathered}\\ \bf \bullet \: The \: Integrate\\ \\\end{gathered}$

SOLUTION :-

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

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

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

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

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

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

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

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

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

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

Answer 2

A N S W E R :

• ${\sf{2\sqrt{x}\:-\:2\:arctan(\sqrt{x})\:+\;C}}$

Given :

• $\displaystyle \sf \large\int\limits^1_0\dfrac{log (x+)}{x} dx$

To find :

• Find The Integrate ?

Solution :

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

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

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

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

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

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

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

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

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

$~~~~~~~$$:\implies{\underline{\boxed{\frak{\pink{2\sqrt{x}\:-\:2\:arctan(\sqrt{x})\:+\;C}}}}}$★

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

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