integral of (x+17)/(x^2+16x+68)

Lösung

\int \frac{x + 17}{x ^{2} + 16 x + 68} d x

\frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2}) + C

Schritte zur Lösung

\int \frac{x + 17}{x ^{2} + 16 x + 68} d x

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 18 u + 85} d u

Vervollständige das Quadrat

u^{2} - 18 u + 85 : (u - 9)^{2} + 4

= \int \frac{u}{( u - 9 ) ^{2} + 4} d u

Wende U-Substitution an

= \int \frac{v + 9}{v ^{2} + 4} d v

Multipliziere aus

\frac{v + 9}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{9}{v ^{2} + 4}

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} + 4} d v + \int \frac{9}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

\int \frac{9}{v ^{2} + 4} d v = \frac{9}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{9}{2} arctan (\frac{v}{2})

Ersetze zurück

= \frac{1}{2} ln (x + 17 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 17 - 9}{2})

Vereinfache

\frac{1}{2} ln (x + 17 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 17 - 9}{2}) : \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2})

= \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2}) + C