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

解

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

\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (2 arctan (\frac{x}{2}) + 2 ln x^{2} + 4) + C

解答ステップ

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

以下の部分分数を得る:

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

= \int \frac{1}{5 ( x - 1 )} + \frac{4 x + 4}{5 ( x ^{2} + 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}{5 ( x - 1 )} d x + \int \frac{4 x + 4}{5 ( x ^{2} + 4 )} d x

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{4 x + 4}{5 ( x ^{2} + 4 )} d x = \frac{1}{5} (2 arctan (\frac{x}{2}) + 2 ln x^{2} + 4)

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (2 arctan (\frac{x}{2}) + 2 ln x^{2} + 4)

定数を解答に追加する

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (2 arctan (\frac{x}{2}) + 2 ln x^{2} + 4) + C