int frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} | vedclass

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(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.Factor out $e^{2025}$ from the numerator and $e^{2026}$ fr

Vedclass · Accepted Answer

$\frac{1}{e} \log_e |e^x + e^{-x}|$

Vedclass · Answer

(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.Factor out $e^{2025}$ from the numerator and $e^{2026}$ from the denominator:$I = \int \frac{e^{2025}(e^x - e^{-x})}{e^{2026}(e^x + e^{-x})} dx = \frac{1}{e} \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.Let $u = e^x + e^{-x}$.Then,the derivative is $du = (e^x - e^{-x}) dx$.Substituting these into the integral:$I = \frac{1}{e} \int \frac{1}{u} du = \frac{1}{e} \ln |u| + C$.Substituting back $u = e^x + e^{-x}$,we get:$I = \frac{1}{e} \ln |e^x + e^{-x}| + C$.

$\int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx = $ . . . . . . + $C$

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