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Beliebt Rechnen >

integral of x/(x^2+2x-1)

Lösung

\int \frac{x}{x ^{2} + 2 x - 1} d x

Lösung

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

Schritte zur Lösung

\int \frac{x}{x ^{2} + 2 x - 1} d x

Faktorisiere

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

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{x}{- x ^{2} - 2 x + 1} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{u - 1}{- u ^{2} + 2} : \frac{u}{- u ^{2} + 2} - \frac{1}{- u ^{2} + 2}

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

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

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

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

Setze in

u = x + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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