integral of (2x^2+9x+2)/((x^2+1)^2)

Lösung

\int \frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} d x

2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

Schritte zur Lösung

\int \frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{2}{x ^{2} + 1} + \frac{9 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{2}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{9}{2 ( x ^{2} + 1 )}

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Füge eine Konstante zur Lösung hinzu

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

\frac{d}{d x} (\frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}})

\int_{0}^{1} tan (\frac{π}{4} x) d x

\int sin^{2} (x) - cos^{2} (x) d x

e^{x} \frac{d y}{d x} + 2 e^{x} y = 1

\frac{d}{d x} (c_{1} e^{- 2 x} + c_{2} x e^{- 2 x})