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

Question

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.

Answer 1

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

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

Answer 2

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!

Answer 3

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