integral from 2 to 1/x of arctan(t)

解

\int_{2}^{\frac{1}{x}} arctan (t) d t

\frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

解答ステップ

\int_{2}^{\frac{1}{x}} arctan (t) d t

部分積分を適用する

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

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

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

限度を計算する:

\frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

= \frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

簡素化

= \frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)