integral of ln(x^2-x-6)

Lösung

\int ln (x^{2} - x - 6) d x

x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1 + C

Schritte zur Lösung

\int ln (x^{2} - x - 6) d x

Vervollständige das Quadrat

x^{2} - x - 6 : (x - \frac{1}{2})^{2} - \frac{25}{4}

= \int ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) d x

Wende U-Substitution an

= \int ln (u^{2} - \frac{25}{4}) d u

Wende die partielle Integration an

= u ln (u^{2} - \frac{25}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} - 25} d u

\int \frac{8 u ^{2}}{4 u ^{2} - 25} d u = - \frac{5}{2} ln \frac{2}{5} u + 1 + \frac{5}{2} ln \frac{2}{5} u - 1 + 2 u

= u ln (u^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} u + 1 + \frac{5}{2} ln \frac{2}{5} u - 1 + 2 u)

Setze in

u = x - \frac{1}{2}

ein

= (x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) + 1 + \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) - 1 + 2 (x - \frac{1}{2}))

Vereinfache

(x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) + 1 + \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) - 1 + 2 (x - \frac{1}{2})) : x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1

= x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1

Füge eine Konstante zur Lösung hinzu

= x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1 + C

x \to 4 + lim (x + x - 4)

l a pl a ce t r an s f or m t - 1

\frac{d}{d x} (\frac{e ^{6 x}}{x ^{2} + 5})

d er i v a t i v e f (t) = (2 t - 1)^{7} + 14 t (2 t - 1)^{6}

y^{^{'}} = (x + y - 4)^{2}, y (0) = 0