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

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

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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!

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$$\int\sqrt{6x+2}\operatorname dx=\frac16\cdot\frac23(6x+2)^{\frac32}+C=\frac19(6x+2)^{\frac32}+C$$, by using the power rule for derivatives and the chain rule.

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