integral from 0 to 0.35 of ln(1+x^2)

Lösung

\int_{0}^{0.35} ln (1 + x^{2}) d x

0.01379 \dots

Schritte zur Lösung

\int_{0}^{0.35} ln (1 + x^{2}) d x

Wende die partielle Integration an

= [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{0.35}

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

= [x ln (1 + x^{2}) - 2 (- arctan (x) + x)]_{0}^{0.35}

Berechne die Grenzen:

0.01379 \dots

= 0.01379 \dots

\int_{0}^{l} x^{2} d x

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

\int_{2}^{5} \frac{1}{x x ^{2} + 1} d x

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

\int_{0}^{1} \frac{x}{x ^{2} - 3 x} d x