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

Lösung

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

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

Dezimale

0.23661 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int_{1}^{e} \frac{1}{x} d x = 1

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

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

Vereinfache

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

\int_{a}^{b} 3 - x^{2} d x

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

\int_{3}^{4} 1

\int_{0}^{9} t d t

\int_{- \frac{π}{4}}^{\frac{π}{4}} 2 sec^{2} (x) d x