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

Lösung

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

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

Schritte zur Lösung

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

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