integral from 2 to 3 of x/(1+x)

Lösung

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

1 - 2 ln (2) + ln (3)

Dezimale

0.71231 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{3}^{4} \frac{u - 1}{u} d u

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int_{3}^{4} 1 - \frac{1}{u} d u

Wende die Summenregel an:

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

= \int_{3}^{4} 1 d u - \int_{3}^{4} \frac{1}{u} d u

\int_{3}^{4} 1 d u = 1

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

= 1 - (2 ln (2) - ln (3))

Vereinfache

= 1 - 2 ln (2) + ln (3)

\int_{0}^{4} sin (x^{2}) d x

\int_{0}^{1} 10 0^{x} d x

\int_{0}^{1} e^{3 + 3 y} - e^{3 y} d y

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

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