• Matéria: Matemática
  • Autor: LucasJairo
  • Perguntado 9 anos atrás

Integral indefinida

 \int\ {} (\frac{(x^2-1)^2}{x^2})  \, dx


Lukyo: ∫ ( x^2 - 1 )^2/x^2 dx

Respostas

respondido por: Lukyo
1
\displaystyle\int\!\frac{(x^2-1)^2}{x^2}\,dx\\\\\\ =\int\!\frac{(x^2)^2-2\cdot x^2\cdot 1+1^2}{x^2}\,dx\\\\\\ =\int\!\frac{x^4-2x^2+1}{x^2}\,dx\\\\\\ =\int\!\frac{x^4}{x^2}\,dx-\int\!\frac{2x^2}{x^2}\,dx+\int\frac{1}{x^2}\,dx

=\displaystyle\int\!x^2\,dx-\int\!2\,dx+\int x^{-2}\,dx\\\\\\ =\frac{x^{2+1}}{2+1}-2x+\frac{x^{-2+1}}{-2+1}+C\\\\\\ =\frac{x^3}{3}-2x+\frac{x^{-1}}{-1}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{(x^2-1)^2}{x^2}\,dx=\frac{x^3}{3}-2x-\frac{1}{x}+C \end{array}}


Bons estudos! :-)

respondido por: CyberKirito
1

 \displaystyle\mathsf{\int\dfrac{(x^2-1)^2}{x^2} \,dx}

\displaystyle\mathsf{\int\frac{x^4-2x^2+1}{x^2} \,dx} \\ \mathsf{=\int ({x}^{2} -  2+  \frac{1}{ {x}^{2} })dx}  \\ \boxed{\boxed{\mathsf{ \frac{1}{3}{x}^{3} - 2x -  \frac{1}{x}  + k}}}

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