Thread: U- sub check.

I have

$\displaystyle \int (3x+1) \sqrt{3x-1}~dx$

$\displaystyle u = 3x-1$

$\displaystyle \frac{du}{du} = 3$

which gives $\displaystyle \frac{1}{3}\int \frac{du}{dx} (u+2)\sqrt{u}~dx = \frac{1}{3}\int (u+2)\sqrt{u}~du $

Is this correct so far?

Originally Posted by Bushy

I have

$\displaystyle \int (3x+1) \sqrt{3x-1}~dx$

$\displaystyle u = 3x-1$

$\displaystyle \frac{du}{du} = 3$

which gives $\displaystyle \frac{1}{3}\int \frac{du}{dx} (u+2)\sqrt{u}~dx = \frac{1}{3}\int (u+2)\sqrt{u}~du $

Is this correct so far?

I'll buy that substitution, although I think you meant $\displaystyle \frac{du}{dx} = 3$ ... continue and finish it.