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

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(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Rewrite the integrand as: $\frac{x - 1}{(x + 1)^3} = \frac{(

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

Vedclass · Answer

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Rewrite the integrand as: $\frac{x - 1}{(x + 1)^3} = \frac{(x + 1) - 2}{(x + 1)^3} = \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3}$. Let $f(x) = \frac{1}{(x + 1)^2}$. Then $f'(x) = \frac{d}{dx} [(x + 1)^{-2}] = -2(x + 1)^{-3} = -\frac{2}{(x + 1)^3}$. Thus,the integral becomes: $\int e^x \left[ \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right] \, dx = \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Substituting $f(x)$ back,we get: $\frac{e^x}{(x + 1)^2} + c$.

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

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