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

Lösung

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

\frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

Dezimale

0.23500 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int_{1}^{2} \frac{x}{x ^{2} + 1} d x = \frac{1}{2} (ln (5) - ln (2))

= ln (2) - \frac{1}{2} (ln (5) - ln (2))

Vereinfache

ln (2) - \frac{1}{2} (ln (5) - ln (2)) : \frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

= \frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

\int_{0}^{1.5} e^{- x} cos (x) d x

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

\int_{0}^{\frac{1}{5}} x ln (5 x) d x

\int_{0}^{2} (\frac{x}{1 + x ^{4}}) d x

\int_{1}^{3} (x + 4) d x