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I have to find this one:

[tex]

\int \frac{dx}{\sqrt{1-e^{2x}}}

[/tex]

Is this right approach?

[tex]

\int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}

[/tex]

Substitution:

[tex]

t = \sqrt{1-e^{2x}}

[/tex]

[tex]

dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx

[/tex]

[tex]

e^{2x} = 1 - t^2\\

[/tex]

[tex]

\int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}} = - \int \frac{dt}{1-t^2} = - \int \frac{1-t}{1-t^2}dt - \int \frac{t}{1-t^2}dt

[/tex]

Substitution:

[tex]

y = 1 - t^2

[/tex]

[tex]

dy = -2t dt

[/tex]

[tex]

z = 1 + t

[/tex]

[tex]

dz = dt

[/tex]

[tex]

... = - \int \frac{dz}{z} - \frac{1}{2} \int \frac{dy}{y} = - \ln \left(1 + \sqrt{1-e^{2x}} \right) - \frac{1}{2} \ln \left(e^{2x} \right) + C

[/tex]

I'm afraid that the first substitution is not ok, but could someone please give me more detailed answer?

Thank you.

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# Homework Help: Integration question - use of substitution

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