integral of (7x^2-4x+2)/(x(x-4)^2)

解

\int \frac{7 x ^{2} - 4 x + 2}{x ( x - 4 ) ^{2}} d x

\frac{1}{8} ln ∣ x ∣ + \frac{55}{8} ln ∣ x - 4 ∣ - \frac{24}{x - 4} - \frac{1}{2 ( x - 4 )} + C

解答ステップ

\int \frac{7 x ^{2} - 4 x + 2}{x ( x - 4 ) ^{2}} d x

拡張

\frac{7 x ^{2} - 4 x + 2}{x ( x - 4 ) ^{2}} : \frac{7 x ^{2}}{x ( x - 4 ) ^{2}} - \frac{4 x}{x ( x - 4 ) ^{2}} + \frac{2}{x ( x - 4 ) ^{2}}

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{7 x ^{2}}{x ( x - 4 ) ^{2}} d x - \int \frac{4 x}{x ( x - 4 ) ^{2}} d x + \int \frac{2}{x ( x - 4 ) ^{2}} d x

\int \frac{7 x ^{2}}{x ( x - 4 ) ^{2}} d x = 7 (ln ∣ x - 4 ∣ - \frac{4}{x - 4})

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

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

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

簡素化

7 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{4}{x - 4}) + 2 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : \frac{1}{8} ln ∣ x ∣ + \frac{55}{8} ln ∣ x - 4 ∣ - \frac{24}{x - 4} - \frac{1}{2 ( x - 4 )}

= \frac{1}{8} ln ∣ x ∣ + \frac{55}{8} ln ∣ x - 4 ∣ - \frac{24}{x - 4} - \frac{1}{2 ( x - 4 )}

定数を解答に追加する

= \frac{1}{8} ln ∣ x ∣ + \frac{55}{8} ln ∣ x - 4 ∣ - \frac{24}{x - 4} - \frac{1}{2 ( x - 4 )} + C