How to evaluate $intfrac{1}{3+4x+4x^2}$?












2












$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21
















2












$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21














2












2








2





$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$





How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 3:35







user587192

















asked Dec 10 '18 at 1:02









YeeetYeeet

133




133












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21


















  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21
















$begingroup$
$$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
$endgroup$
– Will Jagy
Dec 10 '18 at 1:07




$begingroup$
$$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
$endgroup$
– Will Jagy
Dec 10 '18 at 1:07












$begingroup$
Unrelated note: remember to add a $dx$ at the end of your integrals.
$endgroup$
– YiFan
Dec 10 '18 at 1:10




$begingroup$
Unrelated note: remember to add a $dx$ at the end of your integrals.
$endgroup$
– YiFan
Dec 10 '18 at 1:10












$begingroup$
Fixed. Thanks for pointing it out
$endgroup$
– Yeeet
Dec 10 '18 at 1:10




$begingroup$
Fixed. Thanks for pointing it out
$endgroup$
– Yeeet
Dec 10 '18 at 1:10












$begingroup$
so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
$endgroup$
– Yeeet
Dec 10 '18 at 1:20






$begingroup$
so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
$endgroup$
– Yeeet
Dec 10 '18 at 1:20






1




1




$begingroup$
If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
$endgroup$
– YiFan
Dec 10 '18 at 1:21




$begingroup$
If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
$endgroup$
– YiFan
Dec 10 '18 at 1:21










2 Answers
2






active

oldest

votes


















3












$begingroup$

What can be done is a double U-Substitution



We have $$intfrac{1}{3+4x+4x^2}dx$$



By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



$$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    HINT:Complete the squares and use the results



    $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
    and
    $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033282%2fhow-to-evaluate-int-frac134x4x2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      What can be done is a double U-Substitution



      We have $$intfrac{1}{3+4x+4x^2}dx$$



      By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



      Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



      Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
      in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



      $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



      Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



      Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



      so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        What can be done is a double U-Substitution



        We have $$intfrac{1}{3+4x+4x^2}dx$$



        By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



        Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



        Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
        in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



        $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



        Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



        Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



        so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          What can be done is a double U-Substitution



          We have $$intfrac{1}{3+4x+4x^2}dx$$



          By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



          Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



          Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
          in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



          $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



          Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



          Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



          so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






          share|cite|improve this answer









          $endgroup$



          What can be done is a double U-Substitution



          We have $$intfrac{1}{3+4x+4x^2}dx$$



          By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



          Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



          Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
          in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



          $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



          Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



          Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



          so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 1:29









          Aniruddh VenkatesanAniruddh Venkatesan

          156112




          156112























              3












              $begingroup$

              HINT:Complete the squares and use the results



              $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
              and
              $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                HINT:Complete the squares and use the results



                $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                and
                $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  HINT:Complete the squares and use the results



                  $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                  and
                  $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






                  share|cite|improve this answer









                  $endgroup$



                  HINT:Complete the squares and use the results



                  $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                  and
                  $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 1:31









                  Thomas ShelbyThomas Shelby

                  3,4791525




                  3,4791525






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033282%2fhow-to-evaluate-int-frac134x4x2%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more