Question

Evaluate : $\sqrt{ \frac{1 + x}{1 - x} dx \: x≠1}$

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Answer 1

Appropriate Question :-

Evaluate :-

$\rm :\longmapsto\:\displaystyle\int\rm \sqrt{ \frac{1 + x}{1 - x} } \: dx$

$\green{\large\underline{\sf{Solution-}}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \sqrt{ \frac{1 + x}{1 - x} } \: dx$

To evaluate this integral, we first rationalizing the denominator, so on rationalizing the denominator, we get

$\rm \:  =  \: \displaystyle\int\rm \sqrt{ \frac{1 + x}{1 - x} \times \frac{1 + x}{1 + x} } \: dx$

$\rm \:  =  \: \displaystyle\int\rm \sqrt{ \frac{ {(1 + x)}^{2} }{1 - {x}^{2} } } \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{1 + x}{ \sqrt{1 - {x}^{2} } } \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{1}{ \sqrt{1 - {x}^{2} } } \: dx \: + \: \displaystyle\int\rm \frac{x}{ \sqrt{1 - {x}^{2} } } \: dx$

$\rm \:  =  \: {sin}^{ - 1}x - \dfrac{1}{2} \displaystyle\int\rm \frac{ - 2x}{ \sqrt{1 - {x}^{2} } }$

We know,

$\boxed{ \rm{ \: \displaystyle\int\rm \frac{f'(x)}{ \sqrt{f(x)} } = 2 \sqrt{f(x)} + c \: }}$

and

$\red{\rm :\longmapsto\:\dfrac{d}{dx}(1 - {x}^{2}) = - 2x \: }$

So, using this, we get

$\rm \:  =  \: {sin}^{ - 1}x - \dfrac{1}{2} \times 2 \sqrt{1 - {x}^{2} } + c$

$\rm \:  =  \: {sin}^{ - 1}x -\sqrt{1 - {x}^{2} } + c$

Hence,

$\boxed{\displaystyle\int\rm \sqrt{ \frac{1 + x}{1 - x} } \: dx =  \: {sin}^{ - 1}x -\sqrt{1 - {x}^{2} } + c \: }$

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More to Know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

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