Find the derivative of y= u^5/(1+u^3) from 8 to 8-7x

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Homework Statement
use the fundamental theorem of calculus to find the derivative of y= u^5/(1+u^3) from 8 to 8-7x
Relevant Equations
Fundamental Theorem of Calculus
Here is the problem
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Here is my work on it. I thought I did it correct, but again, was told it was wrong.
65888761139__84C90FF9-9408-48B0-82A4-76AC67717D59.JPG
 
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Your answer looks fine to me. My only quibbles are 1) you aren't finding the derivative of ##\frac{u^5}{1 + u^3}## -- the problem asks for the derivative of a specific definite integral of this function; 2) your work is very incomplete - you omitted du, and you don't show what operations you're doing.

If the brain-dead software is flagging your answer as incorrect, it's possible that what you entered isn't syntactically correct -- such as missing parentheses or maybe they were looking for the expanded forms of the numerator and denominator.

What exactly did you enter into the form?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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