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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

x \frac{d y}{d x} = y^{2} - y

\int_{1}^{4} t^{- \frac{3}{2}} d t

x^{^{''}} + 10 x^{^{'}} + 25 = 14 e^{- 5 t}

\frac{d y}{d t} = - 4 (y - 3)

f a c t or x^{4} - 50 x^{2}