Are these legitimate rules/formula for integration without using the substitution method?












0












$begingroup$


I'm talking about:



$int(ax+b)^ndx=frac{(ax+b)^{n+1}}{(n+1)(a)}$



$intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$



$int e^{ax+b}dx=frac{e^{ax+b}}{a}$



$int a^{ax+b}dx=frac{a^{ax+b}}{(ln|a|)(a)}$



Of course, this only works for $(ax+b)$, not for other $f(x)$ with higher exponents like $x^2$ or $x^3$



So for example if write my working as:



begin{align}
int(2x+8)^3dx&=frac{(2x+8)^4}{(4)(2)} \
&=frac{1}{8}(2x+8)^4
end{align}



... based on the rule above



I just want to know if such working/rule applied is actually acceptable without using the substitution method.






integration indefinite-integrals substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
    $endgroup$
    – DavidG
    Dec 31 '18 at 0:35
















0












$begingroup$


I'm talking about:



$int(ax+b)^ndx=frac{(ax+b)^{n+1}}{(n+1)(a)}$



$intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$



$int e^{ax+b}dx=frac{e^{ax+b}}{a}$



$int a^{ax+b}dx=frac{a^{ax+b}}{(ln|a|)(a)}$



Of course, this only works for $(ax+b)$, not for other $f(x)$ with higher exponents like $x^2$ or $x^3$



So for example if write my working as:



begin{align}
int(2x+8)^3dx&=frac{(2x+8)^4}{(4)(2)} \
&=frac{1}{8}(2x+8)^4
end{align}



... based on the rule above



I just want to know if such working/rule applied is actually acceptable without using the substitution method.






integration indefinite-integrals substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
    $endgroup$
    – DavidG
    Dec 31 '18 at 0:35














0












0








0





$begingroup$


I'm talking about:



$int(ax+b)^ndx=frac{(ax+b)^{n+1}}{(n+1)(a)}$



$intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$



$int e^{ax+b}dx=frac{e^{ax+b}}{a}$



$int a^{ax+b}dx=frac{a^{ax+b}}{(ln|a|)(a)}$



Of course, this only works for $(ax+b)$, not for other $f(x)$ with higher exponents like $x^2$ or $x^3$



So for example if write my working as:



begin{align}
int(2x+8)^3dx&=frac{(2x+8)^4}{(4)(2)} \
&=frac{1}{8}(2x+8)^4
end{align}



... based on the rule above



I just want to know if such working/rule applied is actually acceptable without using the substitution method.






integration indefinite-integrals substitution







share|cite|improve this question











$endgroup$




I'm talking about:



$int(ax+b)^ndx=frac{(ax+b)^{n+1}}{(n+1)(a)}$



$intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$



$int e^{ax+b}dx=frac{e^{ax+b}}{a}$



$int a^{ax+b}dx=frac{a^{ax+b}}{(ln|a|)(a)}$



Of course, this only works for $(ax+b)$, not for other $f(x)$ with higher exponents like $x^2$ or $x^3$



So for example if write my working as:



begin{align}
int(2x+8)^3dx&=frac{(2x+8)^4}{(4)(2)} \
&=frac{1}{8}(2x+8)^4
end{align}



... based on the rule above



I just want to know if such working/rule applied is actually acceptable without using the substitution method.






integration indefinite-integrals substitution




integration indefinite-integrals substitution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 1:06









DavidG

1




1










asked Nov 15 '18 at 10:52









HeniasHenias

615




615












  • $begingroup$
    This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
    $endgroup$
    – DavidG
    Dec 31 '18 at 0:35


















  • $begingroup$
    This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
    $endgroup$
    – DavidG
    Dec 31 '18 at 0:35
















$begingroup$
This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
$endgroup$
– DavidG
Dec 31 '18 at 0:35




$begingroup$
This is minor, but in each of your integrals you need to include the constant of integration (generally written as $C$ - as I'm sure you are familiar with).
$endgroup$
– DavidG
Dec 31 '18 at 0:35










2 Answers
2






active

oldest

votes


















0












$begingroup$

Okay I'm going to prove/disprove each one of these bad boys.



#1:
$$I=int(ax+b)^nmathrm{d}x$$
Substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
$$I=frac1aint u^nmathrm{d}u$$
$$I=frac{(ax+b)^{n+1}}{a(n+1)}$$
Correct



#2:
$$I=intfrac{mathrm{d}x}{ax+b}$$
substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
$$I=frac1aintfrac{mathrm{d}u}u$$
$$I=frac1aln|u|$$
$$I=frac1aln|ax+b|, qquad aneq0$$
You forgot the $|$ bars, but otherwise correct



#3:
$$I=int e^{ax+b}mathrm{d}x$$
substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
$$I=frac1aint e^umathrm{d}u$$
$$I=frac1ae^u$$
$$I=frac1ae^{ax+b}$$
Correct



#4:
$$I=int a^{ax+b}mathrm{d}x$$
$$I=int e^{(ax+b)ln a}mathrm{d}x$$
substitution: $u=(ax+b)ln aRightarrow frac{mathrm{d}u}{aln a}=mathrm{d}x$
$$I=frac1{aln a}int e^umathrm{d}u$$
$$I=frac1{aln a}e^u$$
$$I=frac{e^{(ax+b)ln a}}{aln a}$$
$$I=frac{a^{ax+b}}{aln a}$$
$$I=frac{a^{ax+b-1}}{ln a}$$
Basically correct



As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving
$$intfrac{mathrm{d}x}{x}=ln|x|$$
every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The second formula is false, the other formulas are correct.



    The correct second formula reads as follows: for $a ne 0$ we have $intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
      $endgroup$
      – Henias
      Nov 15 '18 at 11:12













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    2 Answers
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    2 Answers
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    active

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    0












    $begingroup$

    Okay I'm going to prove/disprove each one of these bad boys.



    #1:
    $$I=int(ax+b)^nmathrm{d}x$$
    Substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
    $$I=frac1aint u^nmathrm{d}u$$
    $$I=frac{(ax+b)^{n+1}}{a(n+1)}$$
    Correct



    #2:
    $$I=intfrac{mathrm{d}x}{ax+b}$$
    substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
    $$I=frac1aintfrac{mathrm{d}u}u$$
    $$I=frac1aln|u|$$
    $$I=frac1aln|ax+b|, qquad aneq0$$
    You forgot the $|$ bars, but otherwise correct



    #3:
    $$I=int e^{ax+b}mathrm{d}x$$
    substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
    $$I=frac1aint e^umathrm{d}u$$
    $$I=frac1ae^u$$
    $$I=frac1ae^{ax+b}$$
    Correct



    #4:
    $$I=int a^{ax+b}mathrm{d}x$$
    $$I=int e^{(ax+b)ln a}mathrm{d}x$$
    substitution: $u=(ax+b)ln aRightarrow frac{mathrm{d}u}{aln a}=mathrm{d}x$
    $$I=frac1{aln a}int e^umathrm{d}u$$
    $$I=frac1{aln a}e^u$$
    $$I=frac{e^{(ax+b)ln a}}{aln a}$$
    $$I=frac{a^{ax+b}}{aln a}$$
    $$I=frac{a^{ax+b-1}}{ln a}$$
    Basically correct



    As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving
    $$intfrac{mathrm{d}x}{x}=ln|x|$$
    every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Okay I'm going to prove/disprove each one of these bad boys.



      #1:
      $$I=int(ax+b)^nmathrm{d}x$$
      Substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
      $$I=frac1aint u^nmathrm{d}u$$
      $$I=frac{(ax+b)^{n+1}}{a(n+1)}$$
      Correct



      #2:
      $$I=intfrac{mathrm{d}x}{ax+b}$$
      substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
      $$I=frac1aintfrac{mathrm{d}u}u$$
      $$I=frac1aln|u|$$
      $$I=frac1aln|ax+b|, qquad aneq0$$
      You forgot the $|$ bars, but otherwise correct



      #3:
      $$I=int e^{ax+b}mathrm{d}x$$
      substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
      $$I=frac1aint e^umathrm{d}u$$
      $$I=frac1ae^u$$
      $$I=frac1ae^{ax+b}$$
      Correct



      #4:
      $$I=int a^{ax+b}mathrm{d}x$$
      $$I=int e^{(ax+b)ln a}mathrm{d}x$$
      substitution: $u=(ax+b)ln aRightarrow frac{mathrm{d}u}{aln a}=mathrm{d}x$
      $$I=frac1{aln a}int e^umathrm{d}u$$
      $$I=frac1{aln a}e^u$$
      $$I=frac{e^{(ax+b)ln a}}{aln a}$$
      $$I=frac{a^{ax+b}}{aln a}$$
      $$I=frac{a^{ax+b-1}}{ln a}$$
      Basically correct



      As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving
      $$intfrac{mathrm{d}x}{x}=ln|x|$$
      every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Okay I'm going to prove/disprove each one of these bad boys.



        #1:
        $$I=int(ax+b)^nmathrm{d}x$$
        Substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aint u^nmathrm{d}u$$
        $$I=frac{(ax+b)^{n+1}}{a(n+1)}$$
        Correct



        #2:
        $$I=intfrac{mathrm{d}x}{ax+b}$$
        substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aintfrac{mathrm{d}u}u$$
        $$I=frac1aln|u|$$
        $$I=frac1aln|ax+b|, qquad aneq0$$
        You forgot the $|$ bars, but otherwise correct



        #3:
        $$I=int e^{ax+b}mathrm{d}x$$
        substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aint e^umathrm{d}u$$
        $$I=frac1ae^u$$
        $$I=frac1ae^{ax+b}$$
        Correct



        #4:
        $$I=int a^{ax+b}mathrm{d}x$$
        $$I=int e^{(ax+b)ln a}mathrm{d}x$$
        substitution: $u=(ax+b)ln aRightarrow frac{mathrm{d}u}{aln a}=mathrm{d}x$
        $$I=frac1{aln a}int e^umathrm{d}u$$
        $$I=frac1{aln a}e^u$$
        $$I=frac{e^{(ax+b)ln a}}{aln a}$$
        $$I=frac{a^{ax+b}}{aln a}$$
        $$I=frac{a^{ax+b-1}}{ln a}$$
        Basically correct



        As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving
        $$intfrac{mathrm{d}x}{x}=ln|x|$$
        every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.






        share|cite|improve this answer











        $endgroup$



        Okay I'm going to prove/disprove each one of these bad boys.



        #1:
        $$I=int(ax+b)^nmathrm{d}x$$
        Substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aint u^nmathrm{d}u$$
        $$I=frac{(ax+b)^{n+1}}{a(n+1)}$$
        Correct



        #2:
        $$I=intfrac{mathrm{d}x}{ax+b}$$
        substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aintfrac{mathrm{d}u}u$$
        $$I=frac1aln|u|$$
        $$I=frac1aln|ax+b|, qquad aneq0$$
        You forgot the $|$ bars, but otherwise correct



        #3:
        $$I=int e^{ax+b}mathrm{d}x$$
        substitution: $u=ax+bRightarrow frac{mathrm{d}u}a=mathrm{d}x$
        $$I=frac1aint e^umathrm{d}u$$
        $$I=frac1ae^u$$
        $$I=frac1ae^{ax+b}$$
        Correct



        #4:
        $$I=int a^{ax+b}mathrm{d}x$$
        $$I=int e^{(ax+b)ln a}mathrm{d}x$$
        substitution: $u=(ax+b)ln aRightarrow frac{mathrm{d}u}{aln a}=mathrm{d}x$
        $$I=frac1{aln a}int e^umathrm{d}u$$
        $$I=frac1{aln a}e^u$$
        $$I=frac{e^{(ax+b)ln a}}{aln a}$$
        $$I=frac{a^{ax+b}}{aln a}$$
        $$I=frac{a^{ax+b-1}}{ln a}$$
        Basically correct



        As you can see, these all require very simple substitutions to prove. And yes, once you have established a formula you can use that formula and you don't have to go through the work of actually integrating. Example: you don't go through the process of proving
        $$intfrac{mathrm{d}x}{x}=ln|x|$$
        every time you have to integrate it, do you? Of course not. By the same token, the formulas derived above can be used instead of going through the steps of integrating every time.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 16 '18 at 17:27

























        answered Nov 15 '18 at 19:25









        clathratusclathratus

        5,1801438




        5,1801438























            1












            $begingroup$

            The second formula is false, the other formulas are correct.



            The correct second formula reads as follows: for $a ne 0$ we have $intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
              $endgroup$
              – Henias
              Nov 15 '18 at 11:12


















            1












            $begingroup$

            The second formula is false, the other formulas are correct.



            The correct second formula reads as follows: for $a ne 0$ we have $intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
              $endgroup$
              – Henias
              Nov 15 '18 at 11:12
















            1












            1








            1





            $begingroup$

            The second formula is false, the other formulas are correct.



            The correct second formula reads as follows: for $a ne 0$ we have $intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$.






            share|cite|improve this answer









            $endgroup$



            The second formula is false, the other formulas are correct.



            The correct second formula reads as follows: for $a ne 0$ we have $intfrac{1}{ax+b}dx=frac{1}{a}ln(ax+b)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 15 '18 at 11:00









            FredFred

            48.7k11849




            48.7k11849












            • $begingroup$
              Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
              $endgroup$
              – Henias
              Nov 15 '18 at 11:12




















            • $begingroup$
              Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
              $endgroup$
              – Henias
              Nov 15 '18 at 11:12


















            $begingroup$
            Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
            $endgroup$
            – Henias
            Nov 15 '18 at 11:12






            $begingroup$
            Sorry, it's a typo, my bad. So if such formulas are legit, do you know why is it that every answers online uses the substitution method instead of a more straightforward way of getting the answer? socratic.org/questions/what-is-the-antiderivative-of-2x-1-3 quora.com/What-is-the-integrals-of-2x-1-%C2%B2
            $endgroup$
            – Henias
            Nov 15 '18 at 11:12




















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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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