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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x ^{2} + 2 x + 5}{( x + 1 ) ( x ^{2} + 1 )} : \frac{2}{x + 1} + \frac{- x + 3}{x ^{2} + 1}

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

a re a \frac{64}{16 + x ^{2}}, 1

\frac{\partial}{\partial x} (e^{3 x} cos (5 y))

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

\int \frac{3 x ^{2} - 8 x + 13}{( x + 3 ) ( x - 1 ) ^{2}} d x