integral from 0 to 2 of x/(x^2+6x+8)

Lösung

\int_{0}^{2} \frac{x}{x ^{2} + 6 x + 8} d x

\frac{ln ( 3 ) + 3 ln ( \frac{3}{4} )}{2}

Dezimale

0.11778 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{x}{x ^{2} + 6 x + 8} d x

Vervollständige das Quadrat

x^{2} + 6 x + 8 : (x + 3)^{2} - 1

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

Wende U-Substitution an

= \int_{3}^{5} \frac{u - 3}{u ^{2} - 1} d u

Multipliziere aus

\frac{u - 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} - \frac{3}{u ^{2} - 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{3}^{5} \frac{u}{u ^{2} - 1} d u - \int_{3}^{5} \frac{3}{u ^{2} - 1} d u

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

\int_{3}^{5} \frac{3}{u ^{2} - 1} d u = - \frac{3 ln ( \frac{3}{4} )}{2}

= \frac{1}{2} ln (3) - (- \frac{3 ln ( \frac{3}{4} )}{2})

Vereinfache

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

= \frac{ln ( 3 ) + 3 ln ( \frac{3}{4} )}{2}

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