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

Lösung

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

- ln (3) + 2 ln (2)

Dezimale

0.28768 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{( x + 2 ) ( x + 1 )} : - \frac{1}{x + 2} + \frac{1}{x + 1}

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

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

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

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

Vereinfache

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

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

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

\int_{1}^{e} \frac{ln ( 9 x )}{x} d x

\int_{π}^{0} - sin^{2} (t) d t

\int_{8}^{\infty} \frac{2}{x ^{3}} d x