integral from 0 to 1 of ln(3x+1)

Lösung

\int_{0}^{1} ln (3 x + 1) d x

\frac{8}{3} ln (2) - 1

Dezimale

0.84839 \dots

Schritte zur Lösung

\int_{0}^{1} ln (3 x + 1) d x

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{1}^{4} ln (u) d u

Wende die partielle Integration an

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

\int 1 d u = u

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

Berechne die Grenzen:

8 ln (2) - 3

= \frac{1}{3} (8 ln (2) - 3)

Vereinfache

= \frac{8}{3} ln (2) - 1

\int_{x}^{x} e^{\frac{x}{y}} d y

\int_{- 1}^{5} (x^{3} - 3 x + 4) d x

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\int_{0}^{8} t (8 - t) d t

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