integral of (x^2+2x+1)/(x^3-16x)

解

\int \frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} d x

\frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣ + C

解答ステップ

\int \frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} d x

拡張

\frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} : \frac{x ^{2}}{x ^{3} - 16 x} + \frac{2 x}{x ^{3} - 16 x} + \frac{1}{x ^{3} - 16 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{x ^{2}}{x ^{3} - 16 x} d x + \int \frac{2 x}{x ^{3} - 16 x} d x + \int \frac{1}{x ^{3} - 16 x} d x

\int \frac{x ^{2}}{x ^{3} - 16 x} d x = \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣)

\int \frac{2 x}{x ^{3} - 16 x} d x = 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣)

\int \frac{1}{x ^{3} - 16 x} d x = - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) + 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣) - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣

簡素化

\frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) + 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣) - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣ : \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣

定数を解答に追加する

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣ + C