integral from 0 to 1 of ln(r^2+1)r

Lösung

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

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

Dezimale

0.19314 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{2} \frac{ln ( u )}{2} d u

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

\int 1 d u = u

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

Berechne die Grenzen:

2 ln (2) - 1

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

Vereinfache

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

\int_{1}^{2} x^{3} + \frac{6}{x ^{2}} d x

\int_{2}^{4} \frac{1}{x ( ln ( x ) ) ^{9}} d x

\int_{0}^{π} t cos (t) d t

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

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