int_0^1 frac{e^x(x - 1)}{(x + 1)^3} , dx = | vedclass

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(B) ધારો કે $I = \int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx$. અંશને $(x + 1 - 2)$ તરીકે લખી શકાય. $I = \int_0^1 \frac{e^x(x + 1 - 2)}{(x + 1)^3} \, d

Vedclass · Accepted Answer

$\frac{e}{4} - 1$

Vedclass · Answer

(B) ધારો કે $I = \int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx$. અંશને $(x + 1 - 2)$ તરીકે લખી શકાય. $I = \int_0^1 \frac{e^x(x + 1 - 2)}{(x + 1)^3} \, dx = \int_0^1 \frac{e^x}{(x + 1)^2} \, dx - 2 \int_0^1 \frac{e^x}{(x + 1)^3} \, dx$. અહીં આપણે જાણીએ છીએ કે $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$. જો $f(x) = \frac{1}{(x + 1)^2}$ લઈએ,તો $f'(x) = -\frac{2}{(x + 1)^3}$ થાય. તેથી,$\int e^x \left( \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right) \, dx = \frac{e^x}{(x + 1)^2} + C$. નિશ્ચિત સંકલનનું મૂલ્ય મેળવતા: $\left[ \frac{e^x}{(x + 1)^2} \right]_0^1 = \frac{e^1}{(1 + 1)^2} - \frac{e^0}{(0 + 1)^2} = \frac{e}{4} - 1$.

$\int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx = $

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