Question

Calcule a integral por substituição
∫2x²-x/raiz quinta de 4x³-3x²+8

Answer 1

$\displaystyle\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx$

Vamos chamar o radicando do denominador de uma função u:

$u=4x^{3}-3x^{2}+8$

Derivando u:

$du=(4\cdot3x^{3-1}-3\cdot2x^{2-1}+0)dx=(12x^{2}-6x)dx\\\\du=6\cdot(2x^{2}-x)dx~~~\therefore~~~\boxed{\boxed{(2x^{2}-x)dx=\dfrac{1}{6}du}}$

Então:

$\displaystyle\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\int\dfrac{(\frac{1}{6})}{\sqrt[5]{u}}du\\\\\\\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{1}{6}\int\dfrac{1}{u^{1/5}}du\\\\\\\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{1}{6}\int u^{-1/5}du\\\\\\\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{1}{6}\cdot\dfrac{u^{(-1/5)+1}}{(-\frac{1}{5})+1}+constante\\\\\\\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{1}{6}\cdot\dfrac{u^{4/5}}{(\frac{4}{5})}+constante$

$\displaystyle\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{1}{6}\cdot\dfrac{5}{4}u^{4/5}+constante\\\\\\\boxed{\boxed{\int\dfrac{2x^{2}-x}{\sqrt[5]{4x^{3}-3x^{2}+8}}dx=\dfrac{5}{24}(4x^{3}-3x^{2}+8)^{4/5}+constante}}$