integral of (24x^2)/(x^4-20x^2+64)

解

\int \frac{24 x ^{2}}{x ^{4} - 20 x ^{2} + 64} d x

24 (\frac{1}{12} ln ∣ x + 2 ∣ - \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x - 4 ∣) + C

解答ステップ

\int \frac{24 x ^{2}}{x ^{4} - 20 x ^{2} + 64} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \int \frac{x ^{2}}{x ^{4} - 20 x ^{2} + 64} d x

以下の部分分数を得る:

\frac{x ^{2}}{x ^{4} - 20 x ^{2} + 64} : \frac{1}{12 ( x + 2 )} - \frac{1}{12 ( x - 2 )} - \frac{1}{6 ( x + 4 )} + \frac{1}{6 ( x - 4 )}

= 24 \cdot \int \frac{1}{12 ( x + 2 )} - \frac{1}{12 ( x - 2 )} - \frac{1}{6 ( x + 4 )} + \frac{1}{6 ( x - 4 )} 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

= 24 (\int \frac{1}{12 ( x + 2 )} d x - \int \frac{1}{12 ( x - 2 )} d x - \int \frac{1}{6 ( x + 4 )} d x + \int \frac{1}{6 ( x - 4 )} d x)

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

\int \frac{1}{12 ( x - 2 )} d x = \frac{1}{12} ln ∣ x - 2 ∣

\int \frac{1}{6 ( x + 4 )} d x = \frac{1}{6} ln ∣ x + 4 ∣

\int \frac{1}{6 ( x - 4 )} d x = \frac{1}{6} ln ∣ x - 4 ∣

= 24 (\frac{1}{12} ln ∣ x + 2 ∣ - \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x - 4 ∣)

定数を解答に追加する

= 24 (\frac{1}{12} ln ∣ x + 2 ∣ - \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x - 4 ∣) + C