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

Lösung

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

\frac{- i ( Li _{2} ( - 1 i ) - Li _{2} ( 1 i ) ) + i ( Li _{2} ( - 0 i ) - Li _{2} ( 0 i ) )}{2}

Schritte zur Lösung

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

Wende die partielle Integration an

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

\int \frac{arctan ( x )}{x} d x = \frac{1}{2} i (Li_{2} (- i x) - Li_{2} (i x))

= [ln (x) arctan (x) - \frac{1}{2} i (Li_{2} (- i x) - Li_{2} (i x))]_{0}^{1}

Berechne die Grenzen:

\frac{- i ( Li _{2} ( - 1 i ) - Li _{2} ( 1 i ) ) + i ( Li _{2} ( - 0 i ) - Li _{2} ( 0 i ) )}{2}

= \frac{- i ( Li _{2} ( - 1 i ) - Li _{2} ( 1 i ) ) + i ( Li _{2} ( - 0 i ) - Li _{2} ( 0 i ) )}{2}

\int_{1}^{\infty} 2 x^{- 3} d x

\int_{0}^{x} e^{- 2 x} d x

\int_{- 1}^{4} - x^{2} + 3 x + 4 d x

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

\int_{\frac{π}{6}}^{\frac{5 π}{6}} sin (x) d x