integral of (2x^3+x^2-21x+24)/(x^2+2x-8)

Lösung

\int \frac{2 x ^{3} + x ^{2} - 21 x + 24}{x ^{2} + 2 x - 8} d x

x^{2} - 3 x + \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{6} ln \frac{1}{3} (x + 1) - 1 - 4 + C

Schritte zur Lösung

\int \frac{2 x ^{3} + x ^{2} - 21 x + 24}{x ^{2} + 2 x - 8} d x

Vervollständige das Quadrat

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

= \int \frac{2 x ^{3} + x ^{2} - 21 x + 24}{( x + 1 ) ^{2} - 9} d x

Wende U-Substitution an

= \int \frac{- 21 u + 2 ( u - 1 ) ^{3} + ( u - 1 ) ^{2} + 45}{u ^{2} - 9} d u

Lange Division

\frac{- 21 u + 2 ( u - 1 ) ^{3} + ( u - 1 ) ^{2} + 45}{u ^{2} - 9} : 2 u - 5 + \frac{u - 1}{u ^{2} - 9}

= \int 2 u - 5 + \frac{u - 1}{u ^{2} - 9} d u

Wende die Summenregel an:

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

= \int 2 u d u - \int 5 d u + \int \frac{u - 1}{u ^{2} - 9} d u

\int 2 u d u = u^{2}

\int 5 d u = 5 u

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

= u^{2} - 5 u + \frac{1}{2} ln u^{2} - 9 + \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

Setze in

u = x + 1

ein

= (x + 1)^{2} - 5 (x + 1) + \frac{1}{2} ln (x + 1)^{2} - 9 + \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1)

Vereinfache

(x + 1)^{2} - 5 (x + 1) + \frac{1}{2} ln (x + 1)^{2} - 9 + \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1) : x^{2} - 3 x + \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{6} ln \frac{1}{3} (x + 1) - 1 - 4

= x^{2} - 3 x + \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{6} ln \frac{1}{3} (x + 1) - 1 - 4

Füge eine Konstante zur Lösung hinzu

= x^{2} - 3 x + \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{6} ln \frac{1}{3} (x + 1) - 1 - 4 + C

\int 9 x^{2} - 4 d x

t an g e n t f (x) = 4 x^{2} + 2, at x = - 2

\frac{d ^{2} y}{d t ^{2}} = 1 - e^{2 t}

\frac{d}{d s} (ln (3))

\int \frac{1}{9 + 2 x} d x