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#48

x^{\pm x}

Separação

\int x^{\pm x}dx: \int dy\int x^{\pm y}dx, \int dy\int y^{\pm x}dx\\

Primeira parcial

Primeira parcial I

\int x^{\pm y} dx = \dfrac{x^{1\pm y}}{1\pm y} + C

Primeira parcial II

\int y^{\pm x} dx = \pm y^{\pm x}\ln^{-1} y + C

Segunda Parcial

Segunda Parcial I

\int dy\int x^{\pm y}dx = \int dy\left(\dfrac{x^{1\pm y}}{1\pm y} + C\right)

-

= D + Cy + x\int dy\dfrac{x^{\pm y}}{1\pm y}

= D + Cy + x\left[\int x^{\mp y}(1\pm y) dy\right]\circ^{-1}

= D + Cy + x\left[\int x^{\mp y}dy\pm \int yx^{\mp y} dy\right]\circ^{-1}

= D + Cy + x\left[E \mp x^{\mp y}\ln^{-1} x\pm \int yx^{\mp y} dy\right]\circ^{-1}

= D + Cy + x\left[E \mp x^{\mp y}\ln^{-1} x\pm \int yx^{\mp y} dy\right]\circ^{-1}

Segunda Parcial II

\int dy\int y^{\pm x}dx = \int dy(\pm y^{\pm x}\ln^{-1}y + C)\\
= D + Cy \pm \int dy\,y^{\pm x}\ln^{-1}y

Recursos

\int xe^{\pm x}dx = \pm xe^{\pm x} - e^{\pm x} + C\\
= e^{\pm x}(\pm x - 1) + C

Fon­te: cymath

outra coisa

\displaystyle\int\dfrac{e^u}{u'}dx
= \displaystyle\int v' dx
= \int dv
= v + C