Question

Calcule as integrais por substituição:
A) ∫6x³/ x⁴ -5 dx
b)∫7x²-1/raiz cubica de 7x³-3x+5 dx
c)∫2x/ raiz quadrada de x²+4 dx

Answer 1

A)

$\displaystyle\int\dfrac{6x^{3}}{x^{4}-5}dx$

Vamos definir o denominador como uma função u:

$u=x^{4}-5$

Portanto:

$du=4x^{3}dx~~~~\therefore~~~~\dfrac{6}{4}du=\dfrac{6}{4}4x^{3}dx~~~~\therefore~~~~\boxed{\boxed{6x^{3}dx=\dfrac{3}{2}du}}$

Substituindo na integral:

$\displaystyle\int\dfrac{6x^{3}}{x^{4}-5}dx=\int\dfrac{(\frac{3}{2})}{u}du\\\\\\\int\dfrac{6x^{3}}{x^{4}-5}dx=\dfrac{3}{2}\int\dfrac{1}{u}du\\\\\\\int\dfrac{6x^{3}}{x^{4}-5}dx=\dfrac{3}{2}ln|u|+constante\\\\\\\boxed{\boxed{\int\dfrac{6x^{3}}{x^{4}-5}dx=\dfrac{3}{2}ln|x^{4}-5|+constante}}$
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B)

$\displaystyle\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx$

Agora, vamos definir o radicando do denominador como u:

$u=7x^{3}-3x+5$

Logo

$du=3\cdot7x^{2}-3dx=21x^{2}-3dx~~~~\therefore~~~\dfrac{1}{3}du=\dfrac{1}{3}(21x^{2}-3)dx\\\\\\\boxed{\boxed{\dfrac{1}{3}du=7x^{2}-1dx}}$

Substituindo:

$\displaystyle\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx=\int\dfrac{(\frac{1}{3})}{\sqrt[3]{u}}du\\\\\\\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx=\dfrac{1}{3}\int u^{(-1/3)}du\\\\\\\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx=\dfrac{1}{3}\cdot\dfrac{u^{(-[1/3]+1)}}{-(1/3)+1}+constante\\\\\\\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx=\dfrac{1}{3}\dfrac{u^{(2/3)}}{(\frac{2}{3})}+constante\\\\\\\boxed{\boxed{\int\dfrac{7x^{2}-1}{\sqrt[3]{7x^{3}-3x+5}}dx=\dfrac{1}{2}(7x^{3}-3x+5)^{(2/3)}+constante}}$
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C)

$\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx$

Chamando x² + 4 de t:

$t=x^{2}+4~~~~\therefore~~~~\boxed{\boxed{dt=2xdx}}$

Já podemos substituir x² + 4 e 2xdx por t e dt, respectivamente:

$\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=\int\dfrac{1}{\sqrt{t}}dt\\\\\\\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=\int t^{(-1/2)}dt\\\\\\\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=\dfrac{t^{-(1/2)+1}}{-(1/2)+1}+constante\\\\\\\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=\dfrac{t^{(1/2)}}{(1/2)}+constante\\\\\\\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=2\sqrt{t}+constante\\\\\\\boxed{\boxed{\displaystyle\int\dfrac{2x}{\sqrt{x^{2}+4}}dx=2\sqrt{x^{2}+4}+constante}}$