integral from-1/4 to 1/4 of 1/(1-x^2)

解

\int_{- \frac{1}{4}}^{\frac{1}{4}} \frac{1}{1 - x ^{2}} d x

\frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}

十進法表記

0.51082 \dots

解答ステップ

\int_{- \frac{1}{4}}^{\frac{1}{4}} \frac{1}{1 - x ^{2}} d x

共通積分を使用する:

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

= [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{- \frac{1}{4}}^{\frac{1}{4}}

簡素化

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{- \frac{1}{4}}^{\frac{1}{4}} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- \frac{1}{4}}^{\frac{1}{4}}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- \frac{1}{4}}^{\frac{1}{4}}

限度を計算する:

\frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}

= \frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}