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

\int_{1}^{2} \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

\int x^{2} sinh (x) d x

d er i v a t i v e f (x) = 2 x^{- 2} - 3 x^{- \frac{7}{3}}

\int 3 x x - 5 d x

\frac{\partial}{\partial x} (arcsin (x))