When integrated 1/y^2 becomes -1/y why is this?

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when integrated 1/y^2 becomes -1/y why is this?
 
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If you mean integrating with respect to y, try taing the derivative of -1/y.
 
i still don't understand
 
daveb said:
If you mean integrating with respect to y, try taing the derivative of -1/y.

escobar147 said:
i still don't understand
daveb is asking you to find this derivative:
$$\frac{d}{dy}\left(\frac{-1}{y}\right)$$
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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