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I was practicing u-substitution. With the first problem, I was able to rewrite $u^{1/3}$ as the cube root of $u$, but when I did the same approach again with $u^{3/2}$ as the square root of $u^3$, Wolfram Alpha and Symbolab both tell me, that it’s wrong, my integral itself is right only if I don’t take the square root.

I will first show the first integral problem, where I could rewrite the result as the cube root.

$$\int (-5x+3)^{-2/3}dx = \frac{1}{5}\cdot \int u^{-2/3} \\=\frac{3}{5}u^{1/3} =\frac{3\sqrt[3]{u}}{5} =\frac{3\sqrt[3]{-5x+3}}{5}$$

Symbolab tells me, my solution is correct.

Now onto the integral I seem to struggle with:

$$\int \sqrt{3x-4}dx = \frac{1}{3}\cdot \int u^{1/2} =\frac{2}{9}\cdot u^{3/2} =\frac{2\sqrt{u^{3}}}{9} =\frac{2\sqrt{(3x-4)^{3}}}{9}$$

Correct answer:

$$=\frac{2u^{3/2}}{9} =\frac{2(3x-4)^{3/2}}{9}$$

But I’m told it’s wrong, but I did the exact same approach, what am I missing/doing wrong?

I thought, it can be simplified this way. When does this rule not apply and when can I simplify like that? Thank you!

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  • $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$
    – Another User
    Feb 3 at 18:58
  • $\begingroup$ put a dolar $ before and after, when you want too have a math formula $\endgroup$
    – user376343
    Feb 3 at 18:59
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    $\begingroup$ Take the derivative of your answer. Do you get the original integrand? If so, your answer is correct and don't let a computer tell you otherwise. $\endgroup$
    – Matthew Leingang
    Feb 3 at 19:05
  • $\begingroup$ The two expressions are the same, one has a radical sign while the other uses exponents. $\endgroup$
    – bobeyt6
    Feb 3 at 19:07
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    $\begingroup$ Try telling them to assume that $3x > 4$. Computer algebra systems get picky about noninteger powers of numbers that might not be positive. $\endgroup$
    – eyeballfrog
    Feb 3 at 19:12

1 Answer 1

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Personally, I don't think you are doing anything wrong. Your algebraic manipulation is spot on. Even Mathway gives you the correct answer with or without $u$-substitution.

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However, I will also do a manual proof to see (with $u$-substitution, in order to answer the question) to see what is the problem.

Manual Proof:

$\int \sqrt{3x+4} dx = \int (3x+4)^{1/2}$

Substituting $u = 3x + 4$ and finding the derivative of $u$, we find that

$\int \sqrt{3x+4}dx = \frac{1}{3}\int u^{\frac{1}{2}} = \frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})=\frac{2}{9}u^{\frac{3}{2}}=\frac{2(3x+4)^{\frac{3}{2}}} {9}$

Therefore, the OP's answer is correct and it is a Symbolab error.

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  • $\begingroup$ Then maybe an error from Symbolabs side? I don’t know if I should trust myself more than these computations since I’m still lacking in maths. But I am confident in my algebraic manipulation so it just confuses me!! $\endgroup$
    – 434dx
    Feb 3 at 19:26
  • $\begingroup$ @434dx, potentially a Symbolab error. I will do a manual proof as well if it helps. $\endgroup$
    – Yajat Shamji
    Feb 3 at 19:27
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    $\begingroup$ No worries, I took the derivative of my result and I ended up with the function I started with, so it has to be correct. But thank you very much :) $\endgroup$
    – 434dx
    Feb 3 at 19:34

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