integral of (x^2-1)/(2x^2-32)

解

\int \frac{x ^{2} - 1}{2 x ^{2} - 32} d x

\frac{1}{2} x - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

解答ステップ

\int \frac{x ^{2} - 1}{2 x ^{2} - 32} d x

長除法

\frac{x ^{2} - 1}{2 x ^{2} - 32} : \frac{1}{2} + \frac{15}{2 x ^{2} - 32}

= \int \frac{1}{2} + \frac{15}{2 x ^{2} - 32} d x

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} d x + \int \frac{15}{2 x ^{2} - 32} d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{15}{2 x ^{2} - 32} d x = - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

= \frac{1}{2} x - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

定数を解答に追加する

= \frac{1}{2} x - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C