1. ## Integration question

Math1.jpg
Hey there! So, I'm working on an integration problem (shown in the picture above) and I'm not sure if I'm right. The last line is just about my final answer. I'm going to simplify it a little more, but that's pretty much the final product. Did I do anything wrong?

2. Your problem was on line 2. You need to put the numerator in parentheses when "bringing the denominator up"

$\dfrac{x^{1/3}-3}{x^{2/3}} = \left(x^{1/3}-3\right)x^{-2/3} = x^{-1/3}-3x^{-2/3}$

3. The text in the image is as follows:

$\displaystyle{\int\, \dfrac{x^{\frac{1}{3}}\, -\, 3}{x^{\frac{2}{3}}}\, dx}$

$\displaystyle{=\, \int\, x^{\frac{1}{3}}\, -\, 3\, \dot\, x^{-\frac{2}{3}}\, dx}$

$\displaystyle{=\, \frac{1}{\left(\frac{4}{3}\right)}x^{\frac{4}{3}} \, -\, 3x\, \dot\, \frac{1}{\left(\frac{1}{3}\right)}x^{\frac{1}{3}}}$

$\displaystyle{=\, \frac{3x^{\frac{4}{3}}}{4}\, -\, 3x\, \dot\, 3x^{\frac{1}{3}}}$

$\displaystyle{=\, \frac{3x^{\frac{4}{3}}}{4}\, -\, 3x^{\frac{4}{3}}\, +\, C}$

4. It is like that:

$\int \frac{x^{1/3}-3}{x^{2/3}}$ dx = $\int (x^{-1/3}-3x^{-2/3})$ dx=$\frac{3}{2}x^{2/3}-9x^{1/3}+c$

5. Originally Posted by daon2
Your problem was on line 2. You need to put the numerator in parentheses when "bringing the denominator up"

$\dfrac{x^{1/3}-3}{x^{2/3}} = \left(x^{1/3}-3\right)x^{-2/3} = x^{-1/3}-3x^{-2/3}$
Ah, okay! So, I think I fixed it now.

Math2.jpg

Does this seem right?

6. Rather than me confirm, why not take the derivative and see if it is what you started with? I will say that you managed to delete an x from your last step.