integral from 0 to 1 of x/(x^2+9)

Lösung

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

\frac{1}{2} ln (10) - ln (3)

Dezimale

0.05268 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

Wende U-Substitution an

= \int_{9}^{10} \frac{1}{2 u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{9}^{10} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{9}^{10}

Berechne die Grenzen:

ln (10) - 2 ln (3)

= \frac{1}{2} (ln (10) - 2 ln (3))

Vereinfache

= \frac{1}{2} ln (10) - ln (3)

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

\int_{0}^{\infty} t^{2} d t

d er i v a t i v e \frac{1}{4 x - 3}

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\int \frac{tan ( z )}{cos ^{2} ( z )} d z