integral from-1/2 to 0 of (2x+4)/(x^2-1)

解

\int_{- \frac{1}{2}}^{0} \frac{2 x + 4}{x ^{2} - 1} d x

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

十進法表記

- 1.90954 \dots

解答ステップ

\int_{- \frac{1}{2}}^{0} \frac{2 x + 4}{x ^{2} - 1} d x

拡張

\frac{2 x + 4}{x ^{2} - 1} : \frac{2 x}{x ^{2} - 1} + \frac{4}{x ^{2} - 1}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- \frac{1}{2}}^{0} \frac{2 x}{x ^{2} - 1} d x + \int_{- \frac{1}{2}}^{0} \frac{4}{x ^{2} - 1} d x

\int_{- \frac{1}{2}}^{0} \frac{2 x}{x ^{2} - 1} d x = - ln (\frac{3}{4})

\int_{- \frac{1}{2}}^{0} \frac{4}{x ^{2} - 1} d x = - 2 ln (3)

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