Question

Calcule as integrais por substituição
a) ∫x(x²-3) elevado a 5 dx
b) ∫x+1/x²+2x+3 dx

Answer 1

$\displaystyle\int x(x^{2}-3)^{5}dx$

Seja y uma função de x, tal que y = x² - 3. Então:

$dy=2xdx~~~\therefore~~~\boxed{\boxed{xdx=\dfrac{1}{2}dy}}$

Portanto:

$\displaystyle\int x(x^{2}-3)^{5}dx=\int(x^{2}-3)^{5}xdx\\\\\\\int x(x^{2}-3)^{5}dx=\int\dfrac{y^{5}}{2}dy\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{2}\int y^{5}dy\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{2}\cdot\dfrac{y^{6}}{6}+constante\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{12}y^{6}+constante\\\\\\\boxed{\boxed{\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{12}(x^{2}-3)^{6}+constante}}$
____________________________________________

$\displaystyle\int\dfrac{x+1}{x^{2}+2x+3}dx$

Vamos chamar x² + 2x + 3 de v e derivar v:

$dv=2x+2dx~~~\therefore~~~\dfrac{1}{2}dv=\dfrac{1}{2}(2x+2)dx~~~\therefore~~~\boxed{\boxed{x+1dx=\dfrac{1}{2}dv}}$

Substituindo na integral:

$\displaystyle\int\dfrac{x+1}{x^{2}+2x+3}dx=\int\dfrac{(\frac{1}{2})}{v}dv\\\\\\\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}\int\dfrac{1}{v}dv\\\\\\\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}ln|v|+constante\\\\\\\boxed{\boxed{\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}ln(x^{2}+2x+3)+constante}}$

Answer 2

Resposta:

\displaystyle\int x(x^{2}-3)^{5}dx

Seja y uma função de x, tal que y = x² - 3. Então:

dy=2xdx~~~\therefore~~~\boxed{\boxed{xdx=\dfrac{1}{2}dy}}

Portanto:

\displaystyle\int x(x^{2}-3)^{5}dx=\int(x^{2}-3)^{5}xdx\\\\\\\int x(x^{2}-3)^{5}dx=\int\dfrac{y^{5}}{2}dy\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{2}\int y^{5}dy\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{2}\cdot\dfrac{y^{6}}{6}+constante\\\\\\\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{12}y^{6}+constante\\\\\\\boxed{\boxed{\displaystyle\int x(x^{2}-3)^{5}dx=\dfrac{1}{12}(x^{2}-3)^{6}+constante}}

____________________________________________

\displaystyle\int\dfrac{x+1}{x^{2}+2x+3}dx

Vamos chamar x² + 2x + 3 de v e derivar v:

dv=2x+2dx~~~\therefore~~~\dfrac{1}{2}dv=\dfrac{1}{2}(2x+2)dx~~~\therefore~~~\boxed{\boxed{x+1dx=\dfrac{1}{2}dv}}

Substituindo na integral:

\displaystyle\int\dfrac{x+1}{x^{2}+2x+3}dx=\int\dfrac{(\frac{1}{2})}{v}dv\\\\\\\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}\int\dfrac{1}{v}dv\\\\\\\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}ln|v|+constante\\\\\\\boxed{\boxed{\int\dfrac{x+1}{x^{2}+2x+3}dx=\dfrac{1}{2}ln(x^{2}+2x+3)+constante}}

Explicação passo-a-passo: