English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Lösung

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

\frac{π}{2}

Dezimale

1.57079 \dots

Schritte zur Lösung

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

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (1 - x) = x (1 - x)

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

Vereinfache

\frac{x}{x 1 - x} : \frac{x}{1 - x}

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

Vereinfache

\frac{x}{1 - x} : \frac{x}{1 - x}

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= - (- 2 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (v) d v)

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Entferne die Konstante:

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

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

Wende die Summenregel an:

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

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

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= - (- 2 \cdot \frac{1}{2} (\frac{π}{2} + 0))

Vereinfache

- (- 2 \cdot \frac{1}{2} (\frac{π}{2} + 0)) : \frac{π}{2}

= \frac{π}{2}

\int_{1}^{3} (x^{2} + x + 4) d x

\int_{1}^{9} 3 x^{2} + x x + 1 d x

\int_{0}^{1} (x (x^{2} - 1)^{3}) d x

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{5}{4}}} d x

\int_{0}^{\frac{π}{2}} x cos (9 x) d x