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

解

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

\frac{- ln ( t ) - 1}{t} + 1

解答ステップ

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

部分積分を適用する

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

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

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

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

限度を計算する:

\frac{- ln ( t ) - 1}{t} + 1

= \frac{- ln ( t ) - 1}{t} + 1