integral from 0 to 4 of x/(x+9)

Lösung

\int_{0}^{4} \frac{x}{x + 9} d x

4 - 9 (ln (13) - 2 ln (3))

Dezimale

0.69047 \dots

Schritte zur Lösung

\int_{0}^{4} \frac{x}{x + 9} d x

Wende U-Substitution an

= \int_{9}^{13} \frac{u - 9}{u} d u

Multipliziere aus

\frac{u - 9}{u} : 1 - \frac{9}{u}

= \int_{9}^{13} 1 - \frac{9}{u} d u

Wende die Summenregel an:

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

= \int_{9}^{13} 1 d u - \int_{9}^{13} \frac{9}{u} d u

\int_{9}^{13} 1 d u = 4

\int_{9}^{13} \frac{9}{u} d u = 9 (ln (13) - 2 ln (3))

= 4 - 9 (ln (13) - 2 ln (3))

\int_{0}^{10} \frac{5}{10 - x} d x

\int_{- 2}^{1} (x^{4} + x^{3} - 2 x^{2} + 4) d x

\int_{0}^{π} z cos (n z) d z

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\int_{0}^{\frac{3 π}{2}} 1