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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{- t ^{2} - 6 t - 5} d t

\int (\frac{x ^{4} - 3 x x + 2}{x}) d x

\int \frac{- 13 x + 3}{( - 4 x + 2 ) ^{2}} d x

\int (x^{2} - 2 x) e^{- x} d x

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