integral of (x-16)/(x^2+2x-8)

Lösung

\int \frac{x - 16}{x ^{2} + 2 x - 8} d x

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

Schritte zur Lösung

\int \frac{x - 16}{x ^{2} + 2 x - 8} d x

Multipliziere aus

\frac{x - 16}{x ^{2} + 2 x - 8} : \frac{x}{x ^{2} + 2 x - 8} - \frac{16}{x ^{2} + 2 x - 8}

Wende die Summenregel an:

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

= \int \frac{x}{x ^{2} + 2 x - 8} d x - \int \frac{16}{x ^{2} + 2 x - 8} d x

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

\int \frac{16}{x ^{2} + 2 x - 8} d x = - \frac{8}{3} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1)

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (e^{x^{2}} + 2)

\int_{- 9}^{- 8} \frac{1}{x} d x

x^{^{'''}} + x = 0

y^{^{'}} = - sin (π x)

\frac{d}{d x} (ln (- 3 x + 2))