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

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(B) Let $I = \int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx$. We can rewrite the numerator as $(x + 1 - 2)$. $I = \int_0^1 \frac{e^x(x + 1 - 2)}{(x + 1)^

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx$. We can rewrite the numerator as $(x + 1 - 2)$. $I = \int_0^1 \frac{e^x(x + 1 - 2)}{(x + 1)^3} \, dx = \int_0^1 \frac{e^x}{(x + 1)^2} \, dx - 2 \int_0^1 \frac{e^x}{(x + 1)^3} \, dx$. Using integration by parts on the first integral: Let $u = e^x$ and $dv = \frac{1}{(x + 1)^2} \, dx$. Then $du = e^x \, dx$ and $v = -\frac{1}{x + 1}$. $\int_0^1 \frac{e^x}{(x + 1)^2} \, dx = \left[ -\frac{e^x}{x + 1} \right]_0^1 - \int_0^1 \left( -\frac{1}{x + 1} \right) e^x \, dx = \left( -\frac{e}{2} + 1 \right) + \int_0^1 \frac{e^x}{x + 1} \, dx$. This approach is complex. Alternatively,observe the form $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$. Let $f(x) = -\frac{1}{2(x + 1)^2}$. Then $f'(x) = \frac{1}{(x + 1)^3}$. This does not match directly. Let's use $f(x) = \frac{1}{(x + 1)^2}$. Then $f'(x) = -\frac{2}{(x + 1)^3}$. Thus,$\int e^x \left( \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right) \, dx = \frac{e^x}{(x + 1)^2} + C$. Evaluating the definite integral: $\left[ \frac{e^x}{(x + 1)^2} \right]_0^1 = \frac{e^1}{(1 + 1)^2} - \frac{e^0}{(0 + 1)^2} = \frac{e}{4} - 1$.

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

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