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

Lösung

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

b^{2} + 1 - 1

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{b^{2} + 1} \frac{1}{2 u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{b^{2} + 1} \frac{1}{u} d u

Wende die Potenzregel an

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{b^{2} + 1}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{1}^{b^{2} + 1}

Berechne die Grenzen:

2 b^{2} + 1 - 2

= \frac{1}{2} (2 b^{2} + 1 - 2)

Vereinfache

= b^{2} + 1 - 1

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