Question

Me ajudem com essa integral
Integral de x³(x⁴-1)² dx

Answer 1

∫ x³(x⁴-1)² dx

u = x⁴-1 
du = 4x³ dx ⇒ 1/4 du = x³ dx

1/4 ∫ u² du = 1/4 (u³/3) = 1/12 u³ = (x⁴-1)³/12

Answer 2

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===============================================================$\displaystyle\sf{\int x^3(x^4-1)^2~dx=\dfrac{1}{4}\int 4x^3(x^4-1)^2~dx}\\\sf{fac_{\!\!,}a~ u=x^4-1\implies du=4x^3~dx}\\\displaystyle\sf{\dfrac{1}{4}\int 4x^3(x^4-1)^2~dx=\dfrac{1}{4}\int u^2~du=\dfrac{1}{4}\cdot\dfrac{1}{3}u^3+k}\\\sf{\dfrac{1}{4}\int4x^3(x^4-1)^2~dx=\dfrac{1}{12}(x^4-1)^3+k}$