integral of s/(s^2+3s+6)

Lösung

\int \frac{s}{s ^{2} + 3 s + 6} d s

\frac{1}{2} ln 4 s^{2} + 12 s + 24 - \frac{3}{5} arctan (\frac{2}{15} s + \frac{3}{5}) + C

Schritte zur Lösung

\int \frac{s}{s ^{2} + 3 s + 6} d s

Vervollständige das Quadrat

s^{2} + 3 s + 6 : (s + \frac{3}{2})^{2} + \frac{15}{4}

= \int \frac{s}{( s + \frac{3}{2} ) ^{2} + \frac{15}{4}} d s

Wende U-Substitution an

= \int \frac{2 ( 2 u - 3 )}{4 u ^{2} + 15} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{2 u - 3}{4 u ^{2} + 15} d u

Multipliziere aus

\frac{2 u - 3}{4 u ^{2} + 15} : \frac{2 u}{4 u ^{2} + 15} - \frac{3}{4 u ^{2} + 15}

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{2 u}{4 u ^{2} + 15} d u - \int \frac{3}{4 u ^{2} + 15} d u)

\int \frac{2 u}{4 u ^{2} + 15} d u = \frac{1}{4} ln 4 u^{2} + 15

\int \frac{3}{4 u ^{2} + 15} d u = \frac{3}{2 5} arctan (\frac{2}{15} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 15 - \frac{3}{2 5} arctan (\frac{2}{15} u))

Setze in

u = s + \frac{3}{2}

ein

= 2 (\frac{1}{4} ln 4 (s + \frac{3}{2})^{2} + 15 - \frac{3}{2 5} arctan (\frac{2}{15} (s + \frac{3}{2})))

Vereinfache

2 (\frac{1}{4} ln 4 (s + \frac{3}{2})^{2} + 15 - \frac{3}{2 5} arctan (\frac{2}{15} (s + \frac{3}{2}))) : \frac{1}{2} ln 4 s^{2} + 12 s + 24 - \frac{3}{5} arctan (\frac{2}{15} s + \frac{3}{5})

= \frac{1}{2} ln 4 s^{2} + 12 s + 24 - \frac{3}{5} arctan (\frac{2}{15} s + \frac{3}{5})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 s^{2} + 12 s + 24 - \frac{3}{5} arctan (\frac{2}{15} s + \frac{3}{5}) + C

a re a x = - 2 y^{2} + 3 y - 5, x + 3 y = - 13

f (x) = 2 x + 1

f (x) = arcsin (\frac{1}{x})

\int_{0}^{15} \frac{4}{5} x d x

y^{^{'}} + cot (x) y = csc (x)