integral from 0 to 2 of 2tln(1+t)

Lösung

\int_{0}^{2} 2 t ln (1 + t) d t

3 ln (3)

Dezimale

3.29583 \dots

Schritte zur Lösung

\int_{0}^{2} 2 t ln (1 + t) d t

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{2} t ln (1 + t) d t

Wende U-Substitution an

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

Wende die partielle Integration an

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{3}

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3}

Vereinfache

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3} : 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

= 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

Berechne die Grenzen:

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

= 2 \cdot \frac{3}{2} ln (3)

Vereinfache

= 3 ln (3)

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

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