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人気のある微分積分 >

integral of (6x^2)/(x^4-5x^2+4)

解

\int \frac{6 x ^{2}}{x ^{4} - 5 x ^{2} + 4} d x

解

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{6 x ^{2}}{x ^{4} - 5 x ^{2} + 4} d x

定数を除く:

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

= 6 \cdot \int \frac{x ^{2}}{x ^{4} - 5 x ^{2} + 4} d x

以下の部分分数を得る:

\frac{x ^{2}}{x ^{4} - 5 x ^{2} + 4} : \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( x - 2 )}

= 6 \cdot \int \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( x - 2 )} 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

= 6 (\int \frac{1}{6 ( x + 1 )} d x - \int \frac{1}{6 ( x - 1 )} d x - \int \frac{1}{3 ( x + 2 )} d x + \int \frac{1}{3 ( x - 2 )} d x)

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

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

\int \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} ln ∣ x + 2 ∣

\int \frac{1}{3 ( x - 2 )} d x = \frac{1}{3} ln ∣ x - 2 ∣

= 6 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 2 ∣)

簡素化

6 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) : ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣

定数を解答に追加する

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

グラフ

人気の例

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\int 22 tan^{6} (x) d x

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