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

Lösung

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

ln (3) - \frac{1}{2} ln (5) + arctan (3) - \frac{π}{4}

Dezimale

0.75754 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int_{1}^{3} \frac{- x + 1}{x ^{2} + 1} d x = - \frac{1}{2} ln (5) + arctan (3) - \frac{π}{4}

= ln (3) - \frac{1}{2} ln (5) + arctan (3) - \frac{π}{4}

t an g e n t f (x) = 2 x^{2} - x^{4}, (1, 1)

\int_{1}^{8} \frac{ln ( x )}{x ^{2}} d x

n = 1 \sum \infty \frac{6}{n ( n + 1 )}

\frac{d}{d x} (\frac{x}{2} + cos (x))

y d y - x d x = 0