integral from 1 to 3 of 1/((x+1)sqrt(x))

Lösung

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

\frac{π}{6}

Dezimale

0.52359 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= 2 \cdot \int_{1}^{3} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 [arctan (u)]_{1}^{3}

Berechne die Grenzen:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Vereinfache

= \frac{π}{6}

\int_{1}^{8} \frac{12 x}{x ^{2} + 6} d x

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

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