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integration
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\[\int\limits{4x(1-4x)^{3}}\]
This one's a little less intuitive. Let u = 1-4x. Then, you end up with ∫ (1-u)*u^3 du, and you can expand after that - at the end, just plug in the value of u and you're finished.
I had u=1-4x but i dont understand how you got (1-u) x u^3 du
Alright, so we have u = 1-4x, which subs into the integrand as u^3. Then, we have 4x = 1-u, so we can put (1-u) into the equation instead of 4x. Then, we know that du = -4 dx, so dx = -du/4, which you can plug into the integrand as well. I think in total you get -1/4 * ∫(1-u)*u^3 du
thats right! thanks you. i finally understand
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