integral from 1 to e of (x^3+1)/(1+x^2)

Lösung

\int_{1}^{e} \frac{x ^{3} + 1}{1 + x ^{2}} d x

\frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2)) + arctan (e) - \frac{π}{4}

Dezimale

2.91052 \dots

Schritte zur Lösung

\int_{1}^{e} \frac{x ^{3} + 1}{1 + x ^{2}} d x

Multipliziere aus

\frac{x ^{3} + 1}{1 + x ^{2}} : \frac{x ^{3}}{1 + x ^{2}} + \frac{1}{1 + x ^{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_{1}^{e} \frac{x ^{3}}{1 + x ^{2}} d x + \int_{1}^{e} \frac{1}{1 + x ^{2}} d x

\int_{1}^{e} \frac{x ^{3}}{1 + x ^{2}} d x = \frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2))

\int_{1}^{e} \frac{1}{1 + x ^{2}} d x = arctan (e) - \frac{π}{4}

= \frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2)) + arctan (e) - \frac{π}{4}

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