integral of (8x^2+3x+8)/((x^2+1)^2)

Lösung

\int \frac{8 x ^{2} + 3 x + 8}{( x ^{2} + 1 ) ^{2}} d x

8 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

Schritte zur Lösung

\int \frac{8 x ^{2} + 3 x + 8}{( x ^{2} + 1 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{8 x ^{2} + 3 x + 8}{( x ^{2} + 1 ) ^{2}} : \frac{8}{x ^{2} + 1} + \frac{3 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{8}{x ^{2} + 1} + \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8}{x ^{2} + 1} d x + \int \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x)

\int \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{3}{2 ( x ^{2} + 1 )}

= 8 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )}

Füge eine Konstante zur Lösung hinzu

= 8 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

\int \frac{1}{2 x + x ln ( x )} d x

\frac{\partial}{\partial x} (\frac{7 y}{x ^{\frac{1}{2}}})

in v erse l a pl a ce \frac{( e ^{- s} )}{s ^{2}}

\int (5 + 4 x^{2})^{2} d x

\int_{0}^{\frac{π}{8}} cos (2 x) d x