# Why can I rewrite this term as a root, but not the other?

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:

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

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

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!

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Feb 3 at 18:58
• put a dolar $before and after, when you want too have a math formula Feb 3 at 18:59 • 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. Feb 3 at 19:05 • The two expressions are the same, one has a radical sign while the other uses exponents. Feb 3 at 19:07 • Try telling them to assume that$3x > 4\$. Computer algebra systems get picky about noninteger powers of numbers that might not be positive. Feb 3 at 19:12

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.

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.

• 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!! Feb 3 at 19:26
• @434dx, potentially a Symbolab error. I will do a manual proof as well if it helps. Feb 3 at 19:27
• 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 :) Feb 3 at 19:34