integral of (12x+30)/((x^2+2x-8))

Lösung

\int \frac{12 x + 30}{( x ^{2} + 2 x - 8 )} d x

6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

Schritte zur Lösung

\int \frac{12 x + 30}{x ^{2} + 2 x - 8} d x

Vervollständige das Quadrat

x^{2} + 2 x - 8 : (x + 1)^{2} - 9

= \int \frac{12 x + 30}{( x + 1 ) ^{2} - 9} d x

Wende U-Substitution an

= \int \frac{12 u + 18}{u ^{2} - 9} d u

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{2 v + 1}{v ^{2} - 1} : \frac{2 v}{v ^{2} - 1} + \frac{1}{v ^{2} - 1}

Wende die Summenregel an:

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

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

\int \frac{2 v}{v ^{2} - 1} d v = ln v^{2} - 1

\int \frac{1}{v ^{2} - 1} d v = - \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣

= 6 (ln v^{2} - 1 - \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣)

Ersetze zurück

= 6 (ln (\frac{x + 1}{3})^{2} - 1 - \frac{1}{2} ln \frac{x + 1}{3} + 1 + \frac{1}{2} ln \frac{x + 1}{3} - 1)

Vereinfache

6 (ln (\frac{x + 1}{3})^{2} - 1 - \frac{1}{2} ln \frac{x + 1}{3} + 1 + \frac{1}{2} ln \frac{x + 1}{3} - 1) : 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

= 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

Füge eine Konstante zur Lösung hinzu

= 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

\frac{\partial}{\partial y} (x^{3} + 3 x^{2} y - y^{4})

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 15, y^{^{'}} (0) = b

x d y - (2 x + 1) e^{- y} d x = 0

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))