integral of (x^2+1)/(x^2-4)-x

Lösung

\int \frac{x ^{2} + 1}{x ^{2} - 4} - x d x

- \frac{5}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x - \frac{x ^{2}}{2} + C

Schritte zur Lösung

\int \frac{x ^{2} + 1}{x ^{2} - 4} - x d x

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{x ^{2} + 1}{x ^{2} - 4} d x - \int x d x

\int \frac{x ^{2} + 1}{x ^{2} - 4} d x = - \frac{5}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x

\int x d x = \frac{x ^{2}}{2}

= - \frac{5}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x - \frac{x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= - \frac{5}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x - \frac{x ^{2}}{2} + C

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