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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