integral of (x+10)/(sqrt(2-x-3x^2))

Lösung

\int \frac{x + 10}{2 - x - 3 x ^{2}} d x

\frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5})) + C

Schritte zur Lösung

\int \frac{x + 10}{2 - x - 3 x ^{2}} d x

Wende U-Substitution an

= \int \frac{u}{- 3 u ^{2} + 59 u - 288} d u

Vervollständige das Quadrat

- 3 u^{2} + 59 u - 288 : - 3 (u - \frac{59}{6})^{2} + \frac{25}{12}

= \int \frac{u}{- 3 ( u - \frac{59}{6} ) ^{2} + \frac{25}{12}} d u

Wende U-Substitution an

= \int \frac{6 v + 59}{3 - 36 v ^{2} + 25} d v

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{6 v + 59}{- 36 v ^{2} + 25} d v

Multipliziere aus

\frac{6 v + 59}{- 36 v ^{2} + 25} : \frac{6 v}{- 36 v ^{2} + 25} + \frac{59}{- 36 v ^{2} + 25}

= \frac{1}{3} \cdot \int \frac{6 v}{- 36 v ^{2} + 25} + \frac{59}{- 36 v ^{2} + 25} d v

Wende die Summenregel an:

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

= \frac{1}{3} (\int \frac{6 v}{- 36 v ^{2} + 25} d v + \int \frac{59}{- 36 v ^{2} + 25} d v)

\int \frac{6 v}{- 36 v ^{2} + 25} d v = - \frac{1}{6} - 36 v^{2} + 25

\int \frac{59}{- 36 v ^{2} + 25} d v = \frac{59}{6} arcsin (\frac{6}{5} v)

= \frac{1}{3} (- \frac{1}{6} - 36 v^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} v))

Ersetze zurück

= \frac{1}{3} - \frac{1}{6} - 36 (x + 10 - \frac{59}{6})^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} (x + 10 - \frac{59}{6}))

Vereinfache

\frac{1}{3} - \frac{1}{6} - 36 (x + 10 - \frac{59}{6})^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} (x + 10 - \frac{59}{6})) : \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5}))

= \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5})) + C

\frac{d}{d x} (3 x^{2} - 9)

n = 0 \sum \infty \frac{4 ^{n}}{5 ^{n}}

d er i v a t i v e f (x) = (3 x + 70)^{50}

y^{^{''}} + 6.2832 y = 0

n = 0 \sum \infty (\frac{4}{5})^{n}