integral from 4/3 to 4 of 1/(1-x^2)

解

\int_{\frac{4}{3}}^{4} \frac{1}{1 - x ^{2}} d x

\frac{- 2 ln ( 3 ) + ln ( 5 ) - ln ( \frac{7}{3} )}{2}

十進法表記

- 0.71754 \dots

解答ステップ

\int_{\frac{4}{3}}^{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{4}{3}}^{4}

簡素化

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{4}{3}}^{4} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{4}{3}}^{4}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{4}{3}}^{4}

限度を計算する:

\frac{- 2 ln ( 3 ) + ln ( 5 ) - ln ( \frac{7}{3} )}{2}

= \frac{- 2 ln ( 3 ) + ln ( 5 ) - ln ( \frac{7}{3} )}{2}