integral from 1 to sqrt(3 of)arctan(1/r)

解

\int_{1}^{3} arctan (\frac{1}{r}) d r

\frac{2 3 ln ( 2 ) + 2 π - 3 π}{4 3}

十進法表記

0.46807 \dots

解答ステップ

\int_{1}^{3} arctan (\frac{1}{r}) d r

部分積分を適用する

= [r arctan (\frac{1}{r}) - \int - \frac{r}{1 + r ^{2}} d r]_{1}^{3}

\int - \frac{r}{1 + r ^{2}} d r = - \frac{1}{2} ln 1 + r^{2}

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

簡素化

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

= [\frac{1}{2} (2 r arctan (\frac{1}{r}) + ln 1 + r^{2})]_{1}^{3}

限度を計算する:

\frac{2 3 ln ( 2 ) + 2 π - 3 π}{4 3}

= \frac{2 3 ln ( 2 ) + 2 π - 3 π}{4 3}