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Beliebt Rechnen >

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

Lösung

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

Lösung

\frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

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

= \int \frac{x + 16}{( x + 1 ) ^{2} - 9} d x

Wende U-Substitution an

= \int \frac{u + 15}{u ^{2} - 9} d u

Multipliziere aus

\frac{u + 15}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} + \frac{15}{u ^{2} - 9}

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{u}{u ^{2} - 9} d u + \int \frac{15}{u ^{2} - 9} d u

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

\int \frac{15}{u ^{2} - 9} d u = - 5 (\frac{1}{2} ln \frac{u}{3} + 1 - \frac{1}{2} ln \frac{u}{3} - 1)

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

Setze in

u = x + 1

ein

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

Vereinfache

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

= \frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

Graph

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