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Question

Mathematics Question on Integrals of Some Particular Functions

∫dx1−e2x\int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}} is equal to

A

log⁡∣e−x+e−2x−1∣+C\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+C

B

log⁡∣ex+e2x−1∣+C\log |{{e}^{x}}+\sqrt{{{e}^{2x}}-1}|+C

C

−log⁡∣e−x+e−2x−1∣+C-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+C

D

−log⁡∣e−2x+e−2x−1∣+C-\log |{{e}^{-2x}}+\sqrt{{{e}^{-2x}}-1}|+C

Answer

−log⁡∣e−x+e−2x−1∣+C-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+C

Explanation

Solution

Let I=∫dx1−e2x=∫e−xe−2x−1dxI=\int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}}=\int{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}}dx Put t=e−xt={{e}^{-x}}
⇒\Rightarrow dt=−e−xdxdt=-{{e}^{-x}}dx
∴\therefore I=∫−dtt2−1=−log⁡∣t+t2−1∣+cI=\int{-\frac{dt}{\sqrt{{{t}^{2}}-1}}}=-\log |t+\sqrt{{{t}^{2}}-1}|+c
=−log⁡∣e−x+e−2x−1+c=-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}+c