Mathematics

# Evaluate the following integral$\int { \cfrac { { e }^{ 2x } }{ { e }^{ 2x }-2 } } dx\quad$

##### SOLUTION
Let $t={e}^{2x}-2\Rightarrow\,dt=2{e}^{2x}\,dx$

$\Rightarrow\,\dfrac{1}{2}dt={e}^{2x}\,dx$

$\displaystyle\int{\dfrac{{e}^{2x}\,dx}{{e}^{2x}-2}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{t}}$

$=\dfrac{1}{2}\log{\left|t\right|}+c$

$=\dfrac{1}{2}\log{\left|{e}^{2x}-2\right|}+c$ where $t={e}^{2x}-2$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
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