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check below for question:using integration by substitution

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I would set u=x^3. so du=3x^2dx so x^2dx=1/3 * du. So you can rewrite it as 1/3 * (u-5)^0.5du and integrate that more easily. Of course you might wanna pull the 1/3 out and just integrate (u-5)^0.5 du and than just put it back in the end when you resubstitute.
do the working out for me to c plzz

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Remember what I said to you. Look outside the square root, you have x^2. There must be a c such that: cx^2 = du/dx for u = x^3 - 5. That would imply that 3x^2 = cx^2 -> c = 3, so multiply by 3/3.
I end up with: 2/9 * (x^3-5)^(3/2) + C
\[\frac{3}{3} \int\limits_{a}^{b}x^2\sqrt{x^3 - 5}dx = \frac{1}{3} \int\limits_{u}^{v}\sqrt{u}du = \frac{1}{3} \int\limits_{u}^{v}u^{\frac{1}{2}}du\]
wait u is x^3-5 rite? and u is 3x^2?
Yeah, du = 3x^2. That's what I said, I belive.
du = 3x^2dx to be more precise.
@TomLikesPhysics That is correct :-)
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what did i do wrong??
Remember that you have to divide by n+1, so you divide by 3/2. That's the same as multiplying by 2/3, so you have (1/3)(2/3) = 2/9. The rest is ok, well done :-)
You're absolutely correct!

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