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

解

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

\frac{1}{2} (x^{2} - 4 + 4 ln x^{2} - 4) - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

解答ステップ

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

拡張

\frac{x ^{3} + 1}{x ^{2} - 4} : \frac{x ^{3}}{x ^{2} - 4} + \frac{1}{x ^{2} - 4}

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

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

\int \frac{1}{x ^{2} - 4} d x = - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

= \frac{1}{2} (x^{2} - 4 + 4 ln x^{2} - 4) - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

定数を解答に追加する

= \frac{1}{2} (x^{2} - 4 + 4 ln x^{2} - 4) - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C