integral of u/(1+u-2u^2)

Lösung

\int \frac{u}{1 + u - 2 u ^{2}} d u

2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u)) + C

Schritte zur Lösung

\int \frac{u}{1 + u - 2 u ^{2}} d u

Vervollständige das Quadrat

1 + u - 2 u^{2} : - 2 (u - \frac{1}{4})^{2} + \frac{9}{8}

= \int \frac{u}{- 2 ( u - \frac{1}{4} ) ^{2} + \frac{9}{8}} d u

Wende U-Substitution an

= \int \frac{2 ( 4 v + 1 )}{- 16 v ^{2} + 9} d v

Entferne die Konstante:

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

= 2 \cdot \int \frac{4 v + 1}{- 16 v ^{2} + 9} d v

Multipliziere aus

\frac{4 v + 1}{- 16 v ^{2} + 9} : \frac{4 v}{- 16 v ^{2} + 9} + \frac{1}{- 16 v ^{2} + 9}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{4 v}{- 16 v ^{2} + 9} d v + \int \frac{1}{- 16 v ^{2} + 9} d v)

\int \frac{4 v}{- 16 v ^{2} + 9} d v = - \frac{1}{8} ln - 16 v^{2} + 9

\int \frac{1}{- 16 v ^{2} + 9} d v = \frac{1}{24} (ln \frac{4}{3} v + 1 - ln \frac{4}{3} v - 1)

= 2 (- \frac{1}{8} ln - 16 v^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} v + 1 - ln \frac{4}{3} v - 1))

Setze in

v = u - \frac{1}{4}

ein

= 2 (- \frac{1}{8} ln - 16 (u - \frac{1}{4})^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} (u - \frac{1}{4}) + 1 - ln \frac{4}{3} (u - \frac{1}{4}) - 1))

Vereinfache

2 (- \frac{1}{8} ln - 16 (u - \frac{1}{4})^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} (u - \frac{1}{4}) + 1 - ln \frac{4}{3} (u - \frac{1}{4}) - 1)) : 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u))

= 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u))

Füge eine Konstante zur Lösung hinzu

= 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u)) + C

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