integral from 0 to 1 of (4x+4)/(x^2+1)

Lösung

\int_{0}^{1} \frac{4 x + 4}{x ^{2} + 1} d x

2 ln (2) + π

Dezimale

4.52788 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{4 x + 4}{x ^{2} + 1} d x

Multipliziere aus

\frac{4 x + 4}{x ^{2} + 1} : \frac{4 x}{x ^{2} + 1} + \frac{4}{x ^{2} + 1}

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}^{1} \frac{4 x}{x ^{2} + 1} d x + \int_{0}^{1} \frac{4}{x ^{2} + 1} d x

\int_{0}^{1} \frac{4 x}{x ^{2} + 1} d x = 2 ln (2)

\int_{0}^{1} \frac{4}{x ^{2} + 1} d x = π

= 2 ln (2) + π

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