integral from 1 to 3x+2 of x/(1+x^3)

Lösung

\int_{1}^{3 x + 2} \frac{x}{1 + x ^{3}} d x

- \frac{1}{3} ln ∣ 3 x + 3 ∣ + \frac{1}{3} ln (2) + \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{6} ln (\frac{4}{3}) + \frac{1}{3} arctan (3 (2 x + 1)) - \frac{π}{6 3}

Schritte zur Lösung

\int_{1}^{3 x + 2} \frac{x}{1 + x ^{3}} d x

Ermittle den Partialbruch von

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

= \int_{1}^{3 x + 2} - \frac{1}{3 ( x + 1 )} + \frac{x + 1}{3 ( x ^{2} - x + 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_{1}^{3 x + 2} \frac{1}{3 ( x + 1 )} d x + \int_{1}^{3 x + 2} \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x

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

\int_{1}^{3 x + 2} \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{6} (ln 12 x^{2} + 12 x + 4 - ln (\frac{4}{3})) + \frac{1}{6 3} (6 arctan (3 (2 x + 1)) - π)

= - \frac{1}{3} (ln ∣ 3 x + 3 ∣ - ln (2)) + \frac{1}{6} (ln 12 x^{2} + 12 x + 4 - ln (\frac{4}{3})) + \frac{1}{6 3} (6 arctan (3 (2 x + 1)) - π)

Vereinfache

- \frac{1}{3} (ln ∣ 3 x + 3 ∣ - ln (2)) + \frac{1}{6} (ln 12 x^{2} + 12 x + 4 - ln (\frac{4}{3})) + \frac{1}{6 3} (6 arctan (3 (2 x + 1)) - π) : - \frac{1}{3} ln ∣ 3 x + 3 ∣ + \frac{1}{3} ln (2) + \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{6} ln (\frac{4}{3}) + \frac{1}{3} arctan (3 (2 x + 1)) - \frac{π}{6 3}

= - \frac{1}{3} ln ∣ 3 x + 3 ∣ + \frac{1}{3} ln (2) + \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{6} ln (\frac{4}{3}) + \frac{1}{3} arctan (3 (2 x + 1)) - \frac{π}{6 3}

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