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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

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

\int 8 sec^{3} (3 x) d x

y^{^{'}} + y \cdot cos (t) = sin (t) \cdot cos (t)

e x p an d x^{3} (4 - x)

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 2