integral from 0 to 1 of (1+12x)/(1+3x)

Lösung

\int_{0}^{1} \frac{1 + 12 x}{1 + 3 x} d x

4 - 2 ln (2)

Dezimale

2.61370 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1 + 12 x}{1 + 3 x} d x

Lange Division

\frac{1 + 12 x}{1 + 3 x} : 4 - \frac{3}{3 x + 1}

= \int_{0}^{1} 4 - \frac{3}{3 x + 1} d 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} 4 d x - \int_{0}^{1} \frac{3}{3 x + 1} d x

\int_{0}^{1} 4 d x = 4

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

= 4 - 2 ln (2)

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

x \to 0 lim (e^{- 3 x})

d er i v a t i v e y = \frac{9 ( 1 - sin ( x ) )}{2 cos ( x )}

\frac{d}{d x} (x e^{x^{2} y})

\int (sin (x) cos^{5} (x)) d x