integral from 0 to 1 of 1/(xsqrt(1-x))

Lösung

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

divergiert

Schritte zur Lösung

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

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

= - (- 2 \cdot \int_{0}^{1} \frac{1}{- u ^{2} + 1} d u)

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Vereinfache

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

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

Berechne die Grenzen:

divergiert

= divergiert

\int_{1}^{4} (4 x + x) d x

\int_{1}^{5} e^{x} d x

\int_{- 1}^{1} x^{3} 1 - x^{2} d x

\int_{0}^{3} \frac{x}{49 + 5 x} d x

\int_{- 3}^{7} ∣ x ∣ d x