integral of (x-6)/(x^2-16x+65)

Lösung

\int \frac{x - 6}{x ^{2} - 16 x + 65} d x

\frac{1}{2} ln x^{2} - 16 x + 65 + 2 arctan (x - 8) + C

Schritte zur Lösung

\int \frac{x - 6}{x ^{2} - 16 x + 65} d x

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 4 u + 5} d u

Vervollständige das Quadrat

u^{2} - 4 u + 5 : (u - 2)^{2} + 1

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

= \frac{1}{2} ln v^{2} + 1 + 2 arctan (v)

Ersetze zurück

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

Vereinfache

\frac{1}{2} ln (x - 6 - 2)^{2} + 1 + 2 arctan (x - 6 - 2) : \frac{1}{2} ln x^{2} - 16 x + 65 + 2 arctan (x - 8)

= \frac{1}{2} ln x^{2} - 16 x + 65 + 2 arctan (x - 8)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 16 x + 65 + 2 arctan (x - 8) + C

\int 2 sin^{3} (x) cos (x) d x

im pl i c i t \frac{d y}{d x}, x + y = 9

d er i v a t i v e ln (e^{x})

\frac{d}{d x} (3 e^{x} + \frac{8}{3 x})

\int_{0}^{1} π (x)^{2} d x