How to solve this Indefinite Integral step by step?

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Integrate



$$
int
fracsqrt(x^2+1)^5x^6
; dx
$$
Trying this question but unable to solve.
Wolfram does it with hyperbolic substitution, I have been asked to solve this integral without hyperbolic substitution.







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  • 1




    Please use MathJax syntax in order for the question to be more clear!
    – Anastassis Kapetanakis
    Aug 24 at 9:41










  • So what did you try?
    – Matthew Leingang
    Aug 24 at 9:44






  • 1




    Let $x = tan u$ perhaps...
    – Sean Roberson
    Aug 24 at 9:45














up vote
-1
down vote

favorite












Integrate



$$
int
fracsqrt(x^2+1)^5x^6
; dx
$$
Trying this question but unable to solve.
Wolfram does it with hyperbolic substitution, I have been asked to solve this integral without hyperbolic substitution.







share|cite|improve this question


















  • 1




    Please use MathJax syntax in order for the question to be more clear!
    – Anastassis Kapetanakis
    Aug 24 at 9:41










  • So what did you try?
    – Matthew Leingang
    Aug 24 at 9:44






  • 1




    Let $x = tan u$ perhaps...
    – Sean Roberson
    Aug 24 at 9:45












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Integrate



$$
int
fracsqrt(x^2+1)^5x^6
; dx
$$
Trying this question but unable to solve.
Wolfram does it with hyperbolic substitution, I have been asked to solve this integral without hyperbolic substitution.







share|cite|improve this question














Integrate



$$
int
fracsqrt(x^2+1)^5x^6
; dx
$$
Trying this question but unable to solve.
Wolfram does it with hyperbolic substitution, I have been asked to solve this integral without hyperbolic substitution.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 9:56

























asked Aug 24 at 9:35









Chirag Mahajan

156




156







  • 1




    Please use MathJax syntax in order for the question to be more clear!
    – Anastassis Kapetanakis
    Aug 24 at 9:41










  • So what did you try?
    – Matthew Leingang
    Aug 24 at 9:44






  • 1




    Let $x = tan u$ perhaps...
    – Sean Roberson
    Aug 24 at 9:45












  • 1




    Please use MathJax syntax in order for the question to be more clear!
    – Anastassis Kapetanakis
    Aug 24 at 9:41










  • So what did you try?
    – Matthew Leingang
    Aug 24 at 9:44






  • 1




    Let $x = tan u$ perhaps...
    – Sean Roberson
    Aug 24 at 9:45







1




1




Please use MathJax syntax in order for the question to be more clear!
– Anastassis Kapetanakis
Aug 24 at 9:41




Please use MathJax syntax in order for the question to be more clear!
– Anastassis Kapetanakis
Aug 24 at 9:41












So what did you try?
– Matthew Leingang
Aug 24 at 9:44




So what did you try?
– Matthew Leingang
Aug 24 at 9:44




1




1




Let $x = tan u$ perhaps...
– Sean Roberson
Aug 24 at 9:45




Let $x = tan u$ perhaps...
– Sean Roberson
Aug 24 at 9:45










2 Answers
2






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oldest

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up vote
2
down vote



accepted










Let $x=tan u$ therefore $$int
fracsqrt(x^2+1)^5x^6
; dx
=intdfrac1cos^7 udfraccos^6 usin^6 udu=intdfraccos udusin^6 u(1-sin^2 u)$$let $w=sin u$ therefore $$intdfraccos udusin^6 u(1-sin^2 u)=intdfracdww^6(1-w^2)=intdfrac0.51-w+dfrac0.51+w+dfrac1w^6+dfrac1w^4+dfrac1w^2dw\=0.5ln|dfracw+1w-1|-dfrac15w^5-dfrac13w^3-dfrac1w+C$$






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    up vote
    1
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    HINT
    beginalign*
    fracsqrt(x^2+1)^5x^6 = frac1xfracsqrt(x^2+1)^5x^5 = frac1xsqrtfrac(x^2+1)^5x^10 = frac1xsqrtleft(fracx^2+1x^2right)^5 = frac1xsqrtleft(1 + frac1x^2right)^5
    endalign*






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    • Further what is the step ?
      – Chirag Mahajan
      Aug 25 at 2:39










    • Make the substitution $1/x = tan(theta)$.
      – APC89
      Aug 25 at 2:51










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Let $x=tan u$ therefore $$int
    fracsqrt(x^2+1)^5x^6
    ; dx
    =intdfrac1cos^7 udfraccos^6 usin^6 udu=intdfraccos udusin^6 u(1-sin^2 u)$$let $w=sin u$ therefore $$intdfraccos udusin^6 u(1-sin^2 u)=intdfracdww^6(1-w^2)=intdfrac0.51-w+dfrac0.51+w+dfrac1w^6+dfrac1w^4+dfrac1w^2dw\=0.5ln|dfracw+1w-1|-dfrac15w^5-dfrac13w^3-dfrac1w+C$$






    share|cite|improve this answer


























      up vote
      2
      down vote



      accepted










      Let $x=tan u$ therefore $$int
      fracsqrt(x^2+1)^5x^6
      ; dx
      =intdfrac1cos^7 udfraccos^6 usin^6 udu=intdfraccos udusin^6 u(1-sin^2 u)$$let $w=sin u$ therefore $$intdfraccos udusin^6 u(1-sin^2 u)=intdfracdww^6(1-w^2)=intdfrac0.51-w+dfrac0.51+w+dfrac1w^6+dfrac1w^4+dfrac1w^2dw\=0.5ln|dfracw+1w-1|-dfrac15w^5-dfrac13w^3-dfrac1w+C$$






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $x=tan u$ therefore $$int
        fracsqrt(x^2+1)^5x^6
        ; dx
        =intdfrac1cos^7 udfraccos^6 usin^6 udu=intdfraccos udusin^6 u(1-sin^2 u)$$let $w=sin u$ therefore $$intdfraccos udusin^6 u(1-sin^2 u)=intdfracdww^6(1-w^2)=intdfrac0.51-w+dfrac0.51+w+dfrac1w^6+dfrac1w^4+dfrac1w^2dw\=0.5ln|dfracw+1w-1|-dfrac15w^5-dfrac13w^3-dfrac1w+C$$






        share|cite|improve this answer














        Let $x=tan u$ therefore $$int
        fracsqrt(x^2+1)^5x^6
        ; dx
        =intdfrac1cos^7 udfraccos^6 usin^6 udu=intdfraccos udusin^6 u(1-sin^2 u)$$let $w=sin u$ therefore $$intdfraccos udusin^6 u(1-sin^2 u)=intdfracdww^6(1-w^2)=intdfrac0.51-w+dfrac0.51+w+dfrac1w^6+dfrac1w^4+dfrac1w^2dw\=0.5ln|dfracw+1w-1|-dfrac15w^5-dfrac13w^3-dfrac1w+C$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 24 at 10:03

























        answered Aug 24 at 9:58









        Mostafa Ayaz

        10.1k3730




        10.1k3730




















            up vote
            1
            down vote













            HINT
            beginalign*
            fracsqrt(x^2+1)^5x^6 = frac1xfracsqrt(x^2+1)^5x^5 = frac1xsqrtfrac(x^2+1)^5x^10 = frac1xsqrtleft(fracx^2+1x^2right)^5 = frac1xsqrtleft(1 + frac1x^2right)^5
            endalign*






            share|cite|improve this answer




















            • Further what is the step ?
              – Chirag Mahajan
              Aug 25 at 2:39










            • Make the substitution $1/x = tan(theta)$.
              – APC89
              Aug 25 at 2:51














            up vote
            1
            down vote













            HINT
            beginalign*
            fracsqrt(x^2+1)^5x^6 = frac1xfracsqrt(x^2+1)^5x^5 = frac1xsqrtfrac(x^2+1)^5x^10 = frac1xsqrtleft(fracx^2+1x^2right)^5 = frac1xsqrtleft(1 + frac1x^2right)^5
            endalign*






            share|cite|improve this answer




















            • Further what is the step ?
              – Chirag Mahajan
              Aug 25 at 2:39










            • Make the substitution $1/x = tan(theta)$.
              – APC89
              Aug 25 at 2:51












            up vote
            1
            down vote










            up vote
            1
            down vote









            HINT
            beginalign*
            fracsqrt(x^2+1)^5x^6 = frac1xfracsqrt(x^2+1)^5x^5 = frac1xsqrtfrac(x^2+1)^5x^10 = frac1xsqrtleft(fracx^2+1x^2right)^5 = frac1xsqrtleft(1 + frac1x^2right)^5
            endalign*






            share|cite|improve this answer












            HINT
            beginalign*
            fracsqrt(x^2+1)^5x^6 = frac1xfracsqrt(x^2+1)^5x^5 = frac1xsqrtfrac(x^2+1)^5x^10 = frac1xsqrtleft(fracx^2+1x^2right)^5 = frac1xsqrtleft(1 + frac1x^2right)^5
            endalign*







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 24 at 23:09









            APC89

            816215




            816215











            • Further what is the step ?
              – Chirag Mahajan
              Aug 25 at 2:39










            • Make the substitution $1/x = tan(theta)$.
              – APC89
              Aug 25 at 2:51
















            • Further what is the step ?
              – Chirag Mahajan
              Aug 25 at 2:39










            • Make the substitution $1/x = tan(theta)$.
              – APC89
              Aug 25 at 2:51















            Further what is the step ?
            – Chirag Mahajan
            Aug 25 at 2:39




            Further what is the step ?
            – Chirag Mahajan
            Aug 25 at 2:39












            Make the substitution $1/x = tan(theta)$.
            – APC89
            Aug 25 at 2:51




            Make the substitution $1/x = tan(theta)$.
            – APC89
            Aug 25 at 2:51

















             

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