integral of ((6x^2+5x+6))/((x^2+1)^2)

Lösung

\int \frac{( 6 x ^{2} + 5 x + 6 )}{( x ^{2} + 1 ) ^{2}} d x

6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

Schritte zur Lösung

\int \frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} : \frac{6}{x ^{2} + 1} + \frac{5 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{6}{x ^{2} + 1} + \frac{5 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{6}{x ^{2} + 1} d x + \int \frac{5 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

\int \frac{5 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{5}{2 ( x ^{2} + 1 )}

= 6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )}

Füge eine Konstante zur Lösung hinzu

= 6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

\frac{d S}{d r} = k S

\int sin^{3} (u) d u

\frac{d}{d x} ((ln (x))^{- 1})

d er i v a t i v e y = x^{\frac{4}{3}}

in v erse l a pl a ce \frac{1}{s ^{2} - 1}