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

Lösung

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

\frac{1}{2} ln x^{2} + 1

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{1 + x^{2}}

Berechne die Grenzen:

ln x^{2} + 1

= \frac{1}{2} ln x^{2} + 1

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