integral from i to 1 of (z+1)^2

Lösung

\int_{i}^{1} (z + 1)^{2} d z

- i^{2} - i + \frac{- i ^{3} + 1}{3} + 2

Schritte zur Lösung

\int_{i}^{1} (z + 1)^{2} d z

Multipliziere aus

(z + 1)^{2} : z^{2} + 2 z + 1

= \int_{i}^{1} z^{2} + 2 z + 1 d z

Wende die Summenregel an:

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

= \int_{i}^{1} z^{2} d z + \int_{i}^{1} 2 z d z + \int_{i}^{1} 1 d z

\int_{i}^{1} z^{2} d z = \frac{1 - i ^{3}}{3}

\int_{i}^{1} 2 z d z = 1 - i^{2}

\int_{i}^{1} 1 d z = 1 - i

= \frac{1 - i ^{3}}{3} + 1 - i^{2} + 1 - i

Vereinfache

= - i^{2} - i + \frac{- i ^{3} + 1}{3} + 2

\int_{1}^{\infty} \frac{1}{n ^{2} + 1}

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