integral of (40x^2)/(x^4-52x^2+576)

解

\int \frac{40 x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣) + C

解答ステップ

\int \frac{40 x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

定数を除く:

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

= 40 \cdot \int \frac{x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

以下の部分分数を得る:

\frac{x ^{2}}{x ^{4} - 52 x ^{2} + 576} : \frac{1}{10 ( x + 4 )} - \frac{1}{10 ( x - 4 )} - \frac{3}{20 ( x + 6 )} + \frac{3}{20 ( x - 6 )}

= 40 \cdot \int \frac{1}{10 ( x + 4 )} - \frac{1}{10 ( x - 4 )} - \frac{3}{20 ( x + 6 )} + \frac{3}{20 ( x - 6 )} 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

= 40 (\int \frac{1}{10 ( x + 4 )} d x - \int \frac{1}{10 ( x - 4 )} d x - \int \frac{3}{20 ( x + 6 )} d x + \int \frac{3}{20 ( x - 6 )} d x)

\int \frac{1}{10 ( x + 4 )} d x = \frac{1}{10} ln ∣ x + 4 ∣

\int \frac{1}{10 ( x - 4 )} d x = \frac{1}{10} ln ∣ x - 4 ∣

\int \frac{3}{20 ( x + 6 )} d x = \frac{3}{20} ln ∣ x + 6 ∣

\int \frac{3}{20 ( x - 6 )} d x = \frac{3}{20} ln ∣ x - 6 ∣

= 40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣)

定数を解答に追加する

= 40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣) + C