Calcular

Question

<blockquote> <p>Calcular %#% $ #%</p> </blockquote> <p>Primero probé la sustitución $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$ y obtuvo $t=x+2$$$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t-2}+t^2}}dt$$ and use the fact that $$2(\sqrt{e^{t-2}+t^2})'=\frac{1}{\sqrt{e^{t-2}+t^2}} \cdot(e^{t-2}+2t)$% $ $ Using integration by parts we get that we have to calculate (excluding some terms we know) $que es fea entonces el problema inicial. ¿Tienes alguna idea de cómo resolver el problema?</p>

Answer 1

Sugerencia: Que $y = (x+2)e^{-x/2}.$ deben recibir $I=-2\sinh^{-1}{2}$.

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$\displaystyle y = (x+2)e^{-x/2} \implies dy = -\frac{x}{2}e^{-x/2}\,dx$ y $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$

Por lo tanto contamos con $\displaystyle I = \int{-2}^{0}\frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = -2\int{0}^{2}\frac{dy}{\sqrt{1+y^2}} = -2\sinh^{-1}(2)$, afirmó.

Answer 2

Método largo Un método posible es la siguiente. Uso\begin{align} \sum{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^n \, x^n}{4^n} &= \frac{1}{\sqrt{1 + x}} \ \gamma(s,x) &= \int{0}^{x} e^{-t} \, t^{s-1} \, dt \end{align} entonces:\begin{align} I &= \int{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \ &= \int{-2}^{0} \frac{x \, e^{-x/2} \, dx}{\sqrt{1 + \left(\frac{(x+2)^2}{e^{x}}\right)}} \ &= \sum{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int{-2}^{0} e^{-(n+1/2) x} \, x \, (x+2)^{2n} \, dx \ &= \sum{n} \binom{2n}{n} \frac{(-1)^{n}}{4^n} \, e^{2n+1} \, \int{0}^{2} e^{-(n+1/2) t} \, (t-2) \, t^{2n} \, dt \hspace{5mm} t \to x + 2 \ &= \sum{n} \binom{2n}{n} \, \frac{(-1)^{n}}{4^{n}} \, e^{2n+1} \, \left[ \left(\frac{2}{2n+1}\right)^{2n+2} \, \gamma(2n+2, 2n+1) - 2 \, \left(\frac{2}{2n+1}\right)^{2n+1} \, \gamma(2n+1, 2n+1) \right] \ &= 4 \, \sum{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, \left[ \gamma(2n+2, 2n+1) - (2n+1) \, \gamma(2n+1, 2n+1) \right] \ &= - 4 \, \sum{n} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, (2n+1)^{2n+1} \, e^{-(2n+1)} \ &= -4 \, \sum{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n}}{(2n+1)} \ &= - \, \sum{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int{0}^{2} y^{2n} \, dy \ &= -2 \, \int_{0}^{2} \frac{dy}{\sqrt{1+y^2}} \ &= -2 \, \sinh^{-1}(2). \end {Alinee el}

Método corto

\begin{align} I &= \int{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \ &= -2 \int{-2}^{0} \frac{- (x/2) \, e^{-x/2} \, dx}{\sqrt{1 + ((x+2) \, e^{-x/2})^2}} \ \end {Alinee el} que $u = (x+2) \, e^{-x/2}$, $du = -(x/2) \, e^{-x/2} \, dx$, entonces \begin{align} I &= -2 \int{-2}^{0} \frac{du}{\sqrt{1+u^2}} = -2 \left[\sinh^{-1}(u)\right]{-2}^{0} \ &= -2 \, \sinh^{-1}(2). \end {Alinee el}

Es de destacar %#% $ #%