[tex]\bf \displaystyle \int\limits_{-1}^{0}~(4x^6+2x)^3(12x^5+1)\cdot dx\\\\
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u=4x^6+2x\implies \cfrac{du}{dx}=24x^5+2\implies \cfrac{du}{2(12x^5+1)}=dx\\\\
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\displaystyle \int\limits_{-1}^{0}~u^3\underline{(12x^5+1)}\cdot\cfrac{du}{2\underline{(12x^5+1)}}\implies \cfrac{1}{2}\int\limits_{-1}^{0}~u^3\cdot du\\\\
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[tex]\bf \textit{now, we'll change the bounds, using u(x)}
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u(-1)=4(-1)^6+2(-1)\implies u(-1)=2
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u(0)=4(0)^6+2()\implies u(0)=0\\\\
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\displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{2}^{0}\implies \left. \cfrac{u^4}{8} \right]_{2}^{0}\implies [0]-[2]\implies -2[/tex]
