integral from 0 to x of (1-t)/(t^2+15)

Lösung

\int_{0}^{x} \frac{1 - t}{t ^{2} + 15} d t

\frac{1}{15} arctan (\frac{x}{15}) - \frac{1}{2} (ln x^{2} + 15 - ln (15))

Schritte zur Lösung

\int_{0}^{x} \frac{1 - t}{t ^{2} + 15} d t

Multipliziere aus

\frac{1 - t}{t ^{2} + 15} : \frac{1}{t ^{2} + 15} - \frac{t}{t ^{2} + 15}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} \frac{1}{t ^{2} + 15} d t - \int_{0}^{x} \frac{t}{t ^{2} + 15} d t

\int_{0}^{x} \frac{1}{t ^{2} + 15} d t = \frac{1}{15} arctan (\frac{x}{15})

\int_{0}^{x} \frac{t}{t ^{2} + 15} d t = \frac{1}{2} (ln x^{2} + 15 - ln (15))

= \frac{1}{15} arctan (\frac{x}{15}) - \frac{1}{2} (ln x^{2} + 15 - ln (15))

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