integral of (39x^2)/((x+1)(x^2+1))

Lösung

\int \frac{39 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

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

Schritte zur Lösung

\int \frac{39 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 39 \cdot \int \frac{x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Ermittle den Partialbruch von

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

= 39 \cdot \int \frac{1}{2 ( x + 1 )} + \frac{x - 1}{2 ( 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

= 39 (\int \frac{1}{2 ( x + 1 )} d x + \int \frac{x - 1}{2 ( x ^{2} + 1 )} d x)

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

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

= 39 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

Füge eine Konstante zur Lösung hinzu

= 39 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) + C

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