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

解

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

\frac{- ln ( 2 ) - 1}{2} + 1

十進法表記

0.15342 \dots

解答ステップ

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

部分積分を適用する

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

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

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

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 ( t ) - 1}{t}]_{1}^{2}

限度を計算する:

\frac{- ln ( 2 ) - 1}{2} + 1

= \frac{- ln ( 2 ) - 1}{2} + 1