Question:
$ \int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}} $ is equal to
$ \int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}} $ is equal to
Updated On: May 19, 2024
- $ \log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+C $
- $ \log |{{e}^{x}}+\sqrt{{{e}^{2x}}-1}|+C $
- $ -\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+C $
- $ -\log |{{e}^{-2x}}+\sqrt{{{e}^{-2x}}-1}|+C $
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The Correct Option is C
Solution and Explanation
Let $ I=\int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}}=\int{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}}dx $ Put $ t={{e}^{-x}} $
$ \Rightarrow $ $ dt=-{{e}^{-x}}dx $
$ \therefore $ $ I=\int{-\frac{dt}{\sqrt{{{t}^{2}}-1}}}=-\log |t+\sqrt{{{t}^{2}}-1}|+c $
$=-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}+c $
$ \Rightarrow $ $ dt=-{{e}^{-x}}dx $
$ \therefore $ $ I=\int{-\frac{dt}{\sqrt{{{t}^{2}}-1}}}=-\log |t+\sqrt{{{t}^{2}}-1}|+c $
$=-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}+c $
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
