How to Compute the Anti-Derivative of 1/x^3: Explained

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how to compute the ANTI derivative of 1 / x^3
I think I need the formula: f(x) = 1/x^n than f'(x) = -n/x^n+1 but I'm not sure and don't know how to use it.

I know the answer is: -0.5 * x^-2 but have no idea why.

could someone explain this to me please?
 
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The familiar antiderivative formula

\int x^n\, dx = \frac{x^{n+1}}{n+1}+C

also works for negative exponents. Write your fraction as a negative exponent.
 
LCKurtz said:
The familiar antiderivative formula

\int x^n\, dx = \frac{x^{n+1}}{n+1}+C

also works for negative exponents. Write your fraction as a negative exponent.

thanks for your answer.

now how would I put 1/x³ into that formula to get -0.5 * x^-2 ?
 
discy said:
thanks for your answer.

now how would I put 1/x³ into that formula to get -0.5 * x^-2 ?

Write 1/x³ as xn using a negative exponent and use the formua.
 
hm okay. like x^-3. got it.

I guess I should learn this formula, not only because it's a "familiar" one for you guys. But also because for some reason it's not on my formula sheet.

:) tnx for your help.
 
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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