How Do I Find the Antiderivative of x((2x^2) - 2)^3?

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To find the antiderivative of x((2x^2) - 2)^3, the expression can be simplified to 8x(x^2-1). By substituting t = x^2 - 1, the integral becomes ∫4t^3 dt. After integration, the result is t^4, which translates back to (x^2-1)^4. Evaluating this from 0 to 1 yields -1, correcting the initial confusion about the result being 0.
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how do I find to antiderivative of:

x((2x^2) - 2)^3
 
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buffgilville said:
how do I find to antiderivative of:

x((2x^2) - 2)^3

x(2x^2-2)^3=8x(x^2-1). Replace x^2-1 by t, notice that dt=2dx, so

\int x(2x^2-2)^3dx=\int 4 t^3 dt

then replace back

t=x^2-1

ehild
 
Last edited:
from there I was suppose to compute as x is goes from 0 to 1
I got 0, but the answer is -1
 
buffgilville said:
from there I was suppose to compute as x is goes from 0 to 1
I got 0, but the answer is -1
There was a typo in my message. It is

\int x(2x^2-2)^3dx=\int 4 t^3 dt

correctly, which is t^4= (x^2-1)^4.

\big[(x^2-1)^4\big]_0^1=(1-1)^4-(0-1)^4=-1

ehild
 
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