Set u=3x-7u=3x−7
That means that du=3dxdu=3dx. However, you don't have a 3 on the top.
Don't worry. Instead, just multiply the integral by 3/333. That way, you have the 3 sufficient for the substitution.
Your equation becomes:
int (3 * 2)/(3 sqrt(3x-7))dx∫3⋅23√3x−7dx
You can then substitute u into the square root and du for 3dx using the three on the top. You should then get
int 2/(3 sqrtu)∫23√u
Now personally, I like to take the 2/323 out from the integral and write sqrtu√u as u^(-1/2)u−12. Rewriting this, you get
2/3 int u^(-1/2)23∫u−12
Ignore the 2/323 for now and focus on the int u^(-1/2)∫u−12. Add 1 to the power and then multiply by 1/212 since that's the number you get from adding 1 to the power. Simplify and you should have:
(1/3)(u^(1/2))(13)(u12).
Now since u=3x-7u=3x−7, we need to plug that back into the result. Also, the power of 1/212 is the same thing as the square root. Plugging in and writing a square root, we should get:
(1/3)(sqrt(3x-7))(13)(√3x−7)
