Currently trying to tackle MIT's 18.01 OCW 3rd problem set.
Problem 3A-2e: Compute the indefinite integral:
$$ \int{x \over \sqrt{8-2x^2}}dx $$
My solution
Restate the problem as $\int x(8-2x^2)^{-1/2}dx$.
Via substitution or what they call advanced guessing, arrive at a solution of $${-1 \over 2}(8-2x^2)^{1/2} + c$$
OCW's Solution
Guess $(8-2x^2)^{3/2}$ for the non-constant term. Looks good, but it needs a coefficient to get it in the final form: $${-1 \over 6}(8-2x^2)^{3/2} + c$$
My confusion
It appears that if the OCW had restated the problem, it would have done so as $(8-2x^2)^{1/2}$, which is to say that they seem to have addressed only the denominator of the total function. I cannot figure out what happens to the $x$ numerator, nor the status of the radical term as the denominator of a fraction.
I am at such a loss, I began to suspect a typo, but I thought the best thing would be to ask others. The OCW forum is just about useless. Thanks in advance, folks.
You can see that your solution is correct by taking the derivative. The other answer gives the indefinite integral of $x\sqrt{8-2x^2}$. You can see this by taking the derivative of $-\frac16 (8-2x^2)^{3/2}$.
The solution you have now linked to has several errors. First was misreading the problem, or somehow turning $\dfrac{1}{\sqrt u}$ into $u^{1/2}$. Then in the second method, the derivative of $-2x^2$ was incorrectly given as $-4x^2$, then at the next step $x^2$ disappeared to get a result that matches neither the original integrand nor their mistaken integrand.