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

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Question

(A) 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

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

Vedclass · Answer

(A) 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,$dt = ({e^x} - ({ - 1}){e^{ - x}}) \,dx = ({e^x} + {e^{ - x}}) \,dx$. Substituting these into the integral,we get: $I = \int {\frac{1}{t}} \,dt = \log |t| + c$. Replacing $t$ with the original expression: $I = \log |{e^x} - {e^{ - x}}| + c$.

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

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