Question

calcule a integral indefinida∫(x²+4x-²) dx

Answer 1

Propriedades usadas:

${a}^{ - x} = \frac{1}{ {a}^{x} } \\ \\ \displaystyle\int f(x)±g(x)d{x} = \displaystyle\int f(x)d{x}± \displaystyle\int g(x) d{x} \\ \\ \displaystyle\int {x}^{y} d{x} = \frac{ {x}^{y + 1} }{y + 1} \: ,y≠ - 1 \\ \\ \displaystyle\int \frac{1}{ {x}^{y} } d{x } = - \frac{1}{(y - 1) \times {x}^{y - 1} } \: \: \: , \: y≠1$

$\displaystyle\int\left({  {x}^{2} + 4 {x}^{ - 2} }\right) d{x} \\ \displaystyle\int \left( {x}^{2} + 4 \times \frac{1}{ {x}^{2} } \right) d{x } \\ \displaystyle\int\left({  {x}^{2} + \frac{4}{ {x}^{2} } }\right)d{x  } \\ \displaystyle\int {x}^{2} d{x} + \displaystyle\int \frac{4}{ {x}^{2} } d{x} \\ \frac{ {x}^{2 + 1} }{2 + 1} +4 \times \displaystyle\int \frac{1}{ {x}^{2} } d{x  }$

$\frac{ {x}^{3} }{3} + 4 \times \left({  - \frac{1}{x} }\right) \\ \frac{ {x}^{3} }{3} - \frac{ { {4}}}{x} \\ \color{green}\begin{gathered} \boxed{\begin{array}{lr}\large\sf\: \frac{ {x}^{3} }{3} - \frac{4}{x} + C \: ,C \in\mathbb{R} \large \sf \large \sf  \: \end{array}}\end{gathered}$

$\mathcal{Bons \: estudos } \\ \displaystyle\int_ \empty ^ \mathbb{C}     \frac{ - b \: ± \:  \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} d{ t } \boxed{ \boxed{ \mathbb{\displaystyle\Re}\sf{ \gamma  \alpha }\tt{ \pi}\bf{ \nabla}}}$