Question

c) Para complementar as informações, apresente
2x³-1
H(x), uma função primitiva de h(x)=
(2x4-4x)'
sendo H(1)=2₁

Answer 1

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{2x^3-1}{(2x^4-4x)^3}dx\\\underline{\rm fac_{\!\!,}a}\\\sf t=2x^4-4x\\\sf dt=(8x^3-4)dx\\\sf dt=4(2x^3-1)dx\\\sf \dfrac{1}{4}dt=(2x^3-1)dx\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{2x^3-1}{(2x^4-4x)^3}dx=\dfrac{1}{4}\int\dfrac{dt}{t^4}=\dfrac{1}{4}\cdot-\dfrac{1}{3t^3}+k\\\\\displaystyle\sf\int\dfrac{2x^3-1}{(2x^4-4x)^3}dx=\dfrac{1}{4}\int\dfrac{dt}{t^4}=-\dfrac{1}{12t^3}+k\\\\\displaystyle\sf\int\dfrac{2x^3-1}{(2x^4-4x)^3}dx=-\dfrac{1}{12(2x^4-4x)^3}+k\end{array}}$

$\Large\boxed{\begin{array}{l}\sf H(x)=-\dfrac{1}{12(2x^4-4x)^3}+k\\\sf H(1)=-\dfrac{1}{12(2\cdot1^4-4\cdot1)^3}+k\\\\\sf 2=-\dfrac{1}{12(-8)}+k\\\\\sf k+\dfrac{1}{96}=2\\\sf k=2-\dfrac{1}{96}\\\\\sf k=\dfrac{192-1}{96}\\\\\sf k=\dfrac{191}{96}\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{A\,func_{\!\!,}\tilde ao\,pedida\,\acute e}\\\huge\boxed{\boxed{\boxed{\boxed{\sf H(x)=-\dfrac{1}{12(2x^4-4x)^3}+\dfrac{191}{96}}}}}\end{array}}$