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

Lösung

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

\frac{ln ( 3 ) + 2}{4}

Dezimale

0.77465 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{1 + x}{1 + 2 x} : \frac{1}{1 + 2 x} + \frac{x}{1 + 2 x}

Wende die Summenregel an:

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

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

\int_{0}^{1} \frac{1}{1 + 2 x} d x = \frac{1}{2} ln (3)

\int_{0}^{1} \frac{x}{1 + 2 x} d x = \frac{1}{4} (2 - ln (3))

= \frac{1}{2} ln (3) + \frac{1}{4} (2 - ln (3))

Vereinfache

\frac{1}{2} ln (3) + \frac{1}{4} (2 - ln (3)) : \frac{ln ( 3 ) + 2}{4}

= \frac{ln ( 3 ) + 2}{4}

t a y l or \frac{e ^{x}}{1 + x}

\frac{d y}{d x} + a y + b = 0

x \to 3 - lim (\frac{x + 3}{2})

s l o p e (2, 5), (7, 3)

\int (x^{\frac{1}{2}} + 3 x + 2) d x