How do I integrate the function $sqrt{(6x + 2)}$?












2












$begingroup$


How do you integrate $sqrt{(6x + 2)}$?



I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.










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  • 1




    $begingroup$
    Welcome to Math.Stackexchange! What approaches have to tried so far?
    $endgroup$
    – user458276
    Jan 5 at 2:07






  • 3




    $begingroup$
    Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
    $endgroup$
    – JMoravitz
    Jan 5 at 2:08










  • $begingroup$
    I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
    $endgroup$
    – bcloney
    Jan 5 at 2:12






  • 1




    $begingroup$
    Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
    $endgroup$
    – Eevee Trainer
    Jan 5 at 2:17






  • 1




    $begingroup$
    The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
    $endgroup$
    – JMoravitz
    Jan 5 at 2:25


















2












$begingroup$


How do you integrate $sqrt{(6x + 2)}$?



I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to Math.Stackexchange! What approaches have to tried so far?
    $endgroup$
    – user458276
    Jan 5 at 2:07






  • 3




    $begingroup$
    Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
    $endgroup$
    – JMoravitz
    Jan 5 at 2:08










  • $begingroup$
    I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
    $endgroup$
    – bcloney
    Jan 5 at 2:12






  • 1




    $begingroup$
    Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
    $endgroup$
    – Eevee Trainer
    Jan 5 at 2:17






  • 1




    $begingroup$
    The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
    $endgroup$
    – JMoravitz
    Jan 5 at 2:25
















2












2








2


1



$begingroup$


How do you integrate $sqrt{(6x + 2)}$?



I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.










share|cite|improve this question











$endgroup$




How do you integrate $sqrt{(6x + 2)}$?



I've tried to use the following substitutions: let $x = sin(u)$ and $dx = cos(u)$ (along the lines of the Yahoo Answers link). I tried looking for simple examples of integrals with square roots on Yahoo Answers and elsewhere by Googling, but couldn't find any simpler ones, and that substitution got me nowhere.







calculus integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 2:26









Eevee Trainer

5,7471936




5,7471936










asked Jan 5 at 2:06









bcloneybcloney

264




264








  • 1




    $begingroup$
    Welcome to Math.Stackexchange! What approaches have to tried so far?
    $endgroup$
    – user458276
    Jan 5 at 2:07






  • 3




    $begingroup$
    Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
    $endgroup$
    – JMoravitz
    Jan 5 at 2:08










  • $begingroup$
    I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
    $endgroup$
    – bcloney
    Jan 5 at 2:12






  • 1




    $begingroup$
    Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
    $endgroup$
    – Eevee Trainer
    Jan 5 at 2:17






  • 1




    $begingroup$
    The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
    $endgroup$
    – JMoravitz
    Jan 5 at 2:25
















  • 1




    $begingroup$
    Welcome to Math.Stackexchange! What approaches have to tried so far?
    $endgroup$
    – user458276
    Jan 5 at 2:07






  • 3




    $begingroup$
    Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
    $endgroup$
    – JMoravitz
    Jan 5 at 2:08










  • $begingroup$
    I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
    $endgroup$
    – bcloney
    Jan 5 at 2:12






  • 1




    $begingroup$
    Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
    $endgroup$
    – Eevee Trainer
    Jan 5 at 2:17






  • 1




    $begingroup$
    The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
    $endgroup$
    – JMoravitz
    Jan 5 at 2:25










1




1




$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07




$begingroup$
Welcome to Math.Stackexchange! What approaches have to tried so far?
$endgroup$
– user458276
Jan 5 at 2:07




3




3




$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08




$begingroup$
Do you know how to integrate $sqrt{x}$? Have you heard of "u-substitution"?
$endgroup$
– JMoravitz
Jan 5 at 2:08












$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12




$begingroup$
I don't know if my search engine is bad, but I just found out complicated examples. This is a simple one
$endgroup$
– bcloney
Jan 5 at 2:12




1




1




$begingroup$
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17




$begingroup$
Integrating $sqrt x$ is basically the same as integrating any power: $$int x^n dx = frac{1}{n+1}x^{n+1} + C$$ Keep in mind $sqrt x = x^{1/2}$.
$endgroup$
– Eevee Trainer
Jan 5 at 2:17




1




1




$begingroup$
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25






$begingroup$
The technique of "u-substitution" for integration is just the same thing as the technique of using the chain-rule for derivation in reverse. Similarly, finding antiderivatives are essentially analogous to finding derivatives in reverse. Even if it is your first week with integration, I would expect you to have some experience with derivatives and should have seen how to derive $x^{3/2}$. If you know how to derive $x^{3/2}$ then you should be able to figure out how to integrate $x^{1/2}$.
$endgroup$
– JMoravitz
Jan 5 at 2:25












3 Answers
3






active

oldest

votes


















6












$begingroup$

Hints:




  • Make the $u$-substitution $u = 6x+2$.

  • Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have


$$int x^n dx = frac{x^{n+1}}{n+1} + C$$






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    You are looking for:



    $$int(6x+2)^frac12 dx$$



    Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$



    In other words, the above integral is exactly the same as this:
    $$int (u)^frac12 cdot frac 16 du$$
    You can take the constant outside the integral to make this:
    $$frac 16 int u^frac 12 du$$



    And deal with that integral using the normal power rules.



    At the end, don't forget to resubsitute $u=6x+2$ back in!






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Hints:




        • Make the $u$-substitution $u = 6x+2$.

        • Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have


        $$int x^n dx = frac{x^{n+1}}{n+1} + C$$






        share|cite|improve this answer











        $endgroup$


















          6












          $begingroup$

          Hints:




          • Make the $u$-substitution $u = 6x+2$.

          • Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have


          $$int x^n dx = frac{x^{n+1}}{n+1} + C$$






          share|cite|improve this answer











          $endgroup$
















            6












            6








            6





            $begingroup$

            Hints:




            • Make the $u$-substitution $u = 6x+2$.

            • Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have


            $$int x^n dx = frac{x^{n+1}}{n+1} + C$$






            share|cite|improve this answer











            $endgroup$



            Hints:




            • Make the $u$-substitution $u = 6x+2$.

            • Don't forget that $sqrt u = u^{1/2}$ and that, for all $n neq -1$, we have


            $$int x^n dx = frac{x^{n+1}}{n+1} + C$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 2:31

























            answered Jan 5 at 2:07









            Eevee TrainerEevee Trainer

            5,7471936




            5,7471936























                4












                $begingroup$

                You are looking for:



                $$int(6x+2)^frac12 dx$$



                Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$



                In other words, the above integral is exactly the same as this:
                $$int (u)^frac12 cdot frac 16 du$$
                You can take the constant outside the integral to make this:
                $$frac 16 int u^frac 12 du$$



                And deal with that integral using the normal power rules.



                At the end, don't forget to resubsitute $u=6x+2$ back in!






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  You are looking for:



                  $$int(6x+2)^frac12 dx$$



                  Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$



                  In other words, the above integral is exactly the same as this:
                  $$int (u)^frac12 cdot frac 16 du$$
                  You can take the constant outside the integral to make this:
                  $$frac 16 int u^frac 12 du$$



                  And deal with that integral using the normal power rules.



                  At the end, don't forget to resubsitute $u=6x+2$ back in!






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    You are looking for:



                    $$int(6x+2)^frac12 dx$$



                    Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$



                    In other words, the above integral is exactly the same as this:
                    $$int (u)^frac12 cdot frac 16 du$$
                    You can take the constant outside the integral to make this:
                    $$frac 16 int u^frac 12 du$$



                    And deal with that integral using the normal power rules.



                    At the end, don't forget to resubsitute $u=6x+2$ back in!






                    share|cite|improve this answer









                    $endgroup$



                    You are looking for:



                    $$int(6x+2)^frac12 dx$$



                    Notice that if we let $u=6x+2$, $frac{du}{dx}=6$ which leads to $dx=frac16 du$



                    In other words, the above integral is exactly the same as this:
                    $$int (u)^frac12 cdot frac 16 du$$
                    You can take the constant outside the integral to make this:
                    $$frac 16 int u^frac 12 du$$



                    And deal with that integral using the normal power rules.



                    At the end, don't forget to resubsitute $u=6x+2$ back in!







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 5 at 2:24









                    Rhys HughesRhys Hughes

                    5,8061528




                    5,8061528























                        2












                        $begingroup$

                        $$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          $$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            $$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.






                            share|cite|improve this answer









                            $endgroup$



                            $$intsqrt{6x+2}operatorname dx=frac16cdotfrac23(6x+2)^{frac32}+C=frac19(6x+2)^{frac32}+C$$, by using the power rule for derivatives and the chain rule.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 3:20









                            Chris CusterChris Custer

                            11.5k3824




                            11.5k3824






























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