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

解

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

- \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ^{3} - 4 x ^{2}} : - \frac{1}{16 x} - \frac{1}{4 x ^{2}} + \frac{1}{16 ( x - 4 )}

= \int - \frac{1}{16 x} - \frac{1}{4 x ^{2}} + \frac{1}{16 ( 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

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

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

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

\int \frac{1}{16 ( x - 4 )} d x = \frac{1}{16} ln ∣ x - 4 ∣

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

簡素化

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣

定数を解答に追加する

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣ + C