Cómo evaluar la integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$

Question

Evaluar esta integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$

Lo he intentado:

Set : $\sqrt{x+1}=t\mapsto dx=2t\,dt \Longrightarrow I=2\int_1^2\frac{t^4-t^2}{t^4-t^2+1}dt=2\int_1^2(1-\frac{1}{t^4-t^2+1})dt$

Y entonces, puse $(t^2-\frac{1}{2})=\frac{\sqrt{3}}{2}\tan v$ ... Pero no puedo seguir resolviendo esta integral...

Answer 1

SUGERENCIA:

Como $\displaystyle t^4-t^2+1=(t^2+1)^2-3t^2=(t^2+\sqrt3t+1)(t^2-\sqrt3t+1)$

$$\frac1{t^4-t^2+1}=\frac1{2\sqrt3t}\left(\frac1{t^2-\sqrt3t+1}-\frac1{t^2+\sqrt3t+1}\right)$$

Utilizando Sustitución trigonométrica ,

$$t^2-\sqrt3t+1=\frac{4t^2-4\sqrt3t+4}4=\frac{(2t-\sqrt3)^2+1}4$$

Así que listo, $2t-\sqrt3=\tan\phi$ etc.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} I &\equiv \int_{0}^{3}{x\root{x + 1}\,dx \over x^{2} + x + 1} =\int_{0}^{3}x\root{x + 1}\, \pars{{1 \over x + 1/2 - \root{3}\ic/2} - {1 \over x + 1/2 + \root{3}\ic/2}}\, {1 \over \root{3}\ic}\,dx \\[3mm]&={2\root{3} \over 3}\Im\int_{0}^{3} {x\ \overbrace{\root{x + 1}}^{\ds{\equiv\ t}} \over x + 1/2 - \root{3}\ic/2}\,\dd x ={2\root{3} \over 3}\Im\int_{1}^{2} {\pars{t^{2} - 1}t \over \pars{t^{2} - 1} + 1/2 - \root{3}\ic/2}\,2t\,\dd t \\[3mm]&={4\root{3} \over 3}\Im\int_{1}^{2} {\pars{t^{2} - 1}t^{2} \over t^{2} + z}\,\dd t \qquad\mbox{where}\qquad z \equiv -\,\half - {\root{3} \over 2}\,\ic =\expo{4\pi\ic/3} \qquad\qquad\qquad\qquad\qquad\pars{1} \end{align} $z$ es una raíz de $z^{2} + z + 1 = 0$ .

\begin{align} I &={4\root{3} \over 3}\Im\int_{0}^{2} {\bracks{\pars{t^{2} + z} - \pars{z + 1}}\bracks{\pars{t^{2} + z} - z} \over t^{2} + z}\,\dd t \\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2} {\pars{t^{2} + z}^{2} - \pars{2z + 1}\pars{t^{2} + z} +\ \overbrace{z\pars{z + 1}}^{\ds{=\ - 1}} \over t^{2} + z}\,\dd t \\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2} \bracks{\pars{t^{2} + z} - \pars{2z + 1} - {1 \over t^{2} + z}}\,\dd t ={4\root{3} \over 3}\pars{{\root{3} \over 2} - \Im\int_{0}^{2}{\dd t \over t^{2} + z}} \\[3mm]&=2 - {4\root{3} \over 3}\, \Im\pars{\expo{-2\pi\ic/3}\int_{0}^{2\expo{2\pi\ic/3}}{\dd t \over t^{2} + 1}} \end{align}

$$ \mbox{Evaluate the right hand side}:\quad I=2 - {4\root{3} \over 3}\, \Im\bracks{\expo{-2\pi\ic/3}\arctan\pars{2\expo{2\pi\ic/3}}} $$