Estoy bastante seguro de que el método utilizado es sustitución trigonométricas. Pero tengo problemas para configurar y solucionar el problema.
Respuestas
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Felix Marin
Puntos
32763
$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle} \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Con $\ds{x \equiv {1 \over u}}$: \begin{align} \color{#00f}{\large\int{\root{25 - x^{2}} \over x^{4}}\,\dd x} &=\int{\root{25 - 1/u^{2}} \over 1/u^{4}}\, \pars{-\,{\dd u \over u^{2}}} =-\int u\root{25u^{2} - 1}\,\dd u \\[3mm]&=-\,\half\int\root{25u^{2} - 1}\,\dd\pars{u^{2}}=-\,{\pars{25u^{2} - 1}^{3/2} \over 75} =-\,{\bracks{25\pars{1/x}^{2} - 1}^{3/2} \over 75} \\[3mm]&=\color{#00f}{\large -\,{\root{25 - x^{2}} \over 75x^{3}}} + \pars{~\mbox{a constant}~} \end{align}
Wilfred Springer
Puntos
141
Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, lo que conseguimos,
\begin{align} \\\&\int\frac{\sqrt{25-x^2}}{x^4}dx \=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta \=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta \=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}{\sin^4\theta}d\theta \=&\dfrac{1}{25}\int\cot^2\theta \csc^2\theta d\theta \end {Alinee el}
Ahora sustituir $u=\cot \theta$ hasta el final.
