Mathematics

Integrate with respect to x:
$$\dfrac{e^{2x}}{e^{2x}-2}$$.


ANSWER

$$\dfrac{ln(e^{2x}-2)}{2}+C$$


SOLUTION

Consider the given integral.

  $$ I=\int{\dfrac{{{e}^{2X}}}{{{e}^{2X}}-2}}dx $$

 $$ t={{e}^{2X}}-2 $$

 $$ dx=\dfrac{dt}{2{{e}^{2X}}} $$

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

 $$ =\dfrac{1}{2}\ln \left( t \right)+C $$

 $$ =\dfrac{1}{2}\ln \left( {{e}^{2X}}-2 \right)+C $$

Hence, this is the required result

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Single Correct Medium Published on 17th 09, 2020
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