Math, asked by chintumanojb8189, 3 months ago

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

Answers

Answered by shadowsabers03
3

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 \small\text{$e^x,$}

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

Substitute,

\small\text{$\longrightarrow u=e^x$}

\small\text{$\longrightarrow du=e^x\,dx$}

Then (2) becomes,

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

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

Assume,

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

Then (3) becomes,

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

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

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

From (4),

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

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

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

Analysing each coefficient,

  • \small\text{$p+q+r=0$}
  • \small\text{$q-r=0$}
  • \small\text{$-p=1$}

Solving them we get,

  • \small\text{$p=-1$}
  • \small\text{$q=\dfrac{1}{2}$}
  • \small\text{$r=\dfrac{1}{2}$}

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

Undoing substitution \small\text{$u=e^x,$}

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

Then (1) becomes,

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

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

Taking \small\text{$2C_1=C,$}

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

Answered by CoruscatingGarçon
2

Answer:

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

HOPE IT HELPS!

#BE BRAINLY

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