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









    $endgroup$













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      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






























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