integral from 6 to 1/x of arctan(t)

解答

\int_{6}^{\frac{1}{x}} arctan (t) d t

\frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 6 arctan (6) + \frac{1}{2} ln (37)

求解步骤

\int_{6}^{\frac{1}{x}} arctan (t) d t

使用分布积分法

= [t arctan (t) - \int \frac{t}{t ^{2} + 1} d t]_{6}^{\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]_{6}^{\frac{1}{x}}

带入上下限计算后得:

\frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 6 arctan (6) + \frac{1}{2} ln (37)

= \frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 6 arctan (6) + \frac{1}{2} ln (37)

化简

= \frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 6 arctan (6) + \frac{1}{2} ln (37)

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