int {frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} ,d | vedclass

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Question

(B) Let $I = \int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \,dx$. Divide the numerator and denominator by ${e^x}$: $I = \int {\frac{{{e^x} - {e^{ - x}}

Vedclass · Accepted Answer

$\log ({e^{2x}} + 1) - x + c$

Vedclass · Answer

(B) Let $I = \int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \,dx$. Divide the numerator and denominator by ${e^x}$: $I = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \,dx$. Let $t = {e^x} + {e^{ - x}}$. Then $dt = ({e^x} - {e^{ - x}}) \,dx$. Substituting these into the integral: $I = \int {\frac{1}{t}} \,dt = \log |t| + c$. Substituting back $t = {e^x} + {e^{ - x}}$: $I = \log ({e^x} + {e^{ - x}}) + c$. Since ${e^x} + {e^{ - x}} = \frac{{{e^{2x}} + 1}}{{{e^x}}}$,we have: $I = \log \left( \frac{{{e^{2x}} + 1}}{{{e^x}}} \right) + c = \log ({e^{2x}} + 1) - \log ({e^x}) + c = \log ({e^{2x}} + 1) - x + c$.

$\int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \,dx = $

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