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integral of x^3/sqrt1-x^4
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u substitution. put \[u=1-x^4\] \[du =-4x^3 dx\] so \[-\frac{1}{4}du = x^3dx\] and then integrate \[-\frac{1}{4}\int\frac{du}{\sqrt{u}}\] using the power rule
ok thank you i alrdy gave you the good answer. But if you could please check my answer id appreciate it! I got answer of sqrt1-x^4/8
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