Question

Integrate with respect to x

$\frac{ {e}^{2x} - 1}{ {e}^{2x} + 1}$

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Answer 1

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

Given integral is

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

To evaluate this integral, Divide numerator and denominator by e^x, so we get

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

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

$\rm \:  =  \: \displaystyle\int\rm \frac{{e}^{x} - {e}^{ - x}}{{e}^{x} + {e}^{ - x}} \: dx$

Now, to evaluate this, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\:{e}^{x} + {e}^{ - x} = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:({e}^{x} - {e}^{x})dx = dy}$

On substituting all the values in above integral, we get

$\rm \:  =  \: \displaystyle\int\rm \: \frac{dy}{y}$

$\rm \:  =  \: log |y| + c$

$\rm \:  =  \: log\bigg |{e}^{x} + {e}^{ - x}\bigg| + c$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\int\rm \frac{{e}^{2x} - 1}{{e}^{2x} + 1} \: dx = log\bigg |{e}^{x} + {e}^{ - x}\bigg| + c \: }}$

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More to know :-

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

Answer 2

Step-by-step explanation:

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