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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{( 1 + 2 x ) ( 1 + x ^{2} )} : \frac{4}{5 ( 2 x + 1 )} + \frac{- 2 x + 1}{5 ( x ^{2} + 1 )}

= \int \frac{4}{5 ( 2 x + 1 )} + \frac{- 2 x + 1}{5 ( 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{4}{5 ( 2 x + 1 )} d x + \int \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} d x

\int \frac{4}{5 ( 2 x + 1 )} d x = \frac{2}{5} ln ∣ 2 x + 1 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

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