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

Lösung

\int_{0}^{1} \frac{x}{1 - x ^{2}} d x

divergiert

Schritte zur Lösung

\int_{0}^{1} \frac{x}{1 - x ^{2}} d x

Wende U-Substitution an

= \int_{1}^{0} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{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_{0}^{1} \frac{1}{u} d u)

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{0}^{1})

Vereinfache

= \frac{1}{2} [ln ∣ u ∣]_{0}^{1}

Berechne die Grenzen:

divergiert

= divergiert

\int_{0}^{1} \frac{1}{64 - x ^{2}} d x

\int_{2}^{6} (x^{2} - 1) d x

\int_{8}^{x} \frac{1}{t} d t

\int_{0}^{2} (x^{3} + 2 x) d x

\int_{1}^{e} \frac{sin ( ln ( θ ) )}{θ} d θ