Derivative of 1/x^n: Simplifying the Process

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If you want to take the derivative of \frac{1}{x^n} do you turn it into x^-n and go from there or somhow use the fact that the anti derivative of 1/x is ln x?
 
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The fact that "the anti derivative of 1/x is ln x" is irrelevant to the anti-derivative of \frac{1}{x^n}. Write as x-n. Because \frac{1}{x^n}= x^{-n}, the anti-derivative of 1 is the anti-derivative of the other.
 
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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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