Question

integral e^2x +1 /e^2x -1 dx

Answer 1

Given to find,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{e^{2x}+1}{e^{2x}-1}\,dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{e^{2x}-1+2}{e^{2x}-1}\,dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\left(1+\dfrac{2}{e^{2x}-1}\right)\,dx }$

$\small\text{ \displaystyle\longrightarrow I=x+2\int\dfrac{dx}{e^{2x}-1}\quad\dots(1) }$

Consider the integral,

$\small\text{ \displaystyle\longrightarrow I_1=\int\dfrac{dx}{e^{2x}-1} }$

Multiplying both numerator and denominator by $\smalle^x,$

$\small\displaystyle\longrightarrow I_1=\int\dfrac{e^x\,dx}{e^x\left((e^x)^2-1\right)}\quad\dots(2)$

Substitute,

$\small\longrightarrow u=e^x$

$\small\longrightarrow du=e^x\,dx$

Then (2) becomes,

$\small\displaystyle\longrightarrow I_1=\int\dfrac{du}{u\left(u^2-1\right)}$

$\small\displaystyle\longrightarrow I_1=\int\dfrac{du}{u(u-1)(u+1)}\quad\dots(3)$

Assume,

$\small\longrightarrow\dfrac{1}{u(u-1)(u+1)}=\dfrac{p}{u}+\dfrac{q}{u-1}+\dfrac{r}{u+1}\quad\dots(4)$

Then (3) becomes,

$\small\displaystyle\longrightarrow I_1=\int\left(\dfrac{p}{u}+\dfrac{q}{u-1}+\dfrac{r}{u+1}\right)\,dx$

$\small\longrightarrow I_1=p\log|u|+q\log|u-1|+r\log|u+1|+C_1$

$\small\longrightarrow I_1=\log\left|u^p(u-1)^q(u+1)^r\right|+C_1\quad\dots(5)$

From (4),

$\small\longrightarrow\dfrac{1}{u(u-1)(u+1)}=\dfrac{p(u-1)(u+1)+qu(u+1)+ru(u-1)}{u(u-1)(u+1)}$

$\small\longrightarrow1=p(u^2-1)+q(u^2+u)+r(u^2-u)$

$\small\longrightarrow (p+q+r)u^2+(q-r)u-p=1$

Analysing each coefficient,

• $\smallp+q+r=0$

• $\smallq-r=0$

• $\small-p=1$

Solving them we get,

• $\smallp=-1$

• $\smallq=\dfrac{1}{2}$

• $\smallr=\dfrac{1}{2}$

Then (5) becomes,

$\small\text{ \longrightarrow I_1=\log\left|u^{-1}(u-1)^{\frac{1}{2}}(u+1)^{\frac{1}{2}}\right|+C_1 }$

$\small\longrightarrow I_1=\dfrac{1}{2}\log\left|u^2-1\right|-\log|u|+C_1$

Undoing substitution $\smallu=e^x,$

$\small\longrightarrow I_1=\dfrac{1}{2}\log\left|e^{2x}-1\right|-x+C_1$

Then (1) becomes,

$\small\displaystyle\longrightarrow I=x+2\left(\dfrac{1}{2}\log\left|e^{2x}-1\right|-x+C_1\right)$

$\small\displaystyle\longrightarrow I=\log\left|e^{2x}-1\right|-x+2C_1$

Taking $\small2C_1=C,$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\log\left|e^{2x}-1\right|-x+C}} }$

Answer 2

Answer:

The value of integral is log|e^2x - 1| - x + c

HOPE IT HELPS!

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