integral from 0 to 2 of 1/(x^2+4x+3)

Lösung

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 3} d x

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

Dezimale

0.29389 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 3} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} + 4 x + 3} : \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x + 3 )}

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

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

\int_{0}^{2} \frac{1}{2 ( x + 3 )} d x = \frac{1}{2} (ln (5) - ln (3))

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

Vereinfache

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

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

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

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