Question

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

Answer 1

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

We will solve the above problem of integration by using substitution method :-

⟹$let \: \sqrt{x} = t$

⟹$(x = {t}^{2} )$

⟹${x}^{ \frac{1}{2} } = t$

⟹$\dfrac{1}{2} {x}^{ \frac{ - 1}{2} } dx = dt$

⟹$\dfrac{dx}{2 \sqrt{x} } = dt$

⟹$dx = 2 \sqrt{x} dt$

⟹$dx = 2 t \: dt$

Now , we get :-

⟹$\displaystyle \int \dfrac{2 {t}^{2} \: dt}{1 + {t}^{2} }\:dx$

⟹$2\displaystyle \int \dfrac{ {t}^{2} + 1 - 1\: }{1 + {t}^{2} }\:dt$

⟹$2\displaystyle \int \dfrac{ {t}^{2} + 1 \: }{1 + {t}^{2} }\:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt$

⟹$2\displaystyle \int1 \:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt$

⟹$2t - 2 {\tan}^{ - 1} (t) + c$

where c = constant

Now put t = √x

$2\sqrt{x} - 2 {\tan}^{ - 1} (\sqrt{x} ) + c$

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

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Answer 2

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 :-

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

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

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

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

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

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

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

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

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

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