1. ## Find the integral

Hello, i need help with this problem, here's what i'm doing.
(6x-7)^1/2 dx

would it not be 3/2(6x-7)^3/2

2. No.

To make it more clear for you, sub u = 6x-7, du = 6 dx. Therefore:

$\int (6x-7)^{\frac{1}{2}}~dx = \frac{1}{6} \int (u)^{\frac{1}{2}}~du = \frac{u^{\frac{3}{2}}}{9} + C$

3. Hello, plstevens!

Hello, i need help with this problem.

Here's what i'm doing: . $\int (6x-7)^{\frac{1}{2}}\,dx$

would it not be: . $\frac{3}{2}(6x-7)^{\frac{3}{2}}$ . . . . no
At least three errors . . .

Let $u \:= \:6x-7 \quad\Rightarrow\quad du \:=\:6\,dx \quad\Rightarrow\quad dx \:=\:\tfrac{1}{6}\,du$

Substitute: . $\int u^{\frac{1}{2}}\left(\tfrac{1}{6}\,du\right) \;=\;\tfrac{1}{6}\int u^{\frac{1}{2}}\,du \;=\; \frac{\frac{1}{6}}{\frac{3}{2}}\,u^{\frac{3}{2}} + C \;=\;\tfrac{1}{9}u^{\frac{3}{2}} + C$

Back-substitute: . $\tfrac{1}{9}(6x-7)^{\frac{3}{2}} + C$

4. ## Solution

Well, you would start by letting u = 6x - 7
Take the derivative of both sides to get:
du = 6dx; Solve for dx to get: dx = du/6. Now substitute this into the original integral to get:

∫[(u)^(1/2)]du/6

Note, u^1/2 is the same as sqrt(u)

Now, because 1/6 is a constant you can put it in front of the integral.

(1/6)∫[u^(1/2)]du

By using the reverse power rule, the integration would result in:
(1/6)*(2/3)*(u^3/2) + C, for some constant C.

Reduce the fractions and substitute the u value back in to yield:

(1/9)*[(6x-7)^3/2] + C.

Note: Since the integral is indefinite, you must include the constant C. Teachers will always take off points on tests if you forget this. I hope this helps. Also, you could take the derivative of this to check that you get (6x-7)^(1/2)