integral of ((x^2+5x+4))/(x^2-2x+1)

解

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

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

解答ステップ

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

拡張

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

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

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

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

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

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 5 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{4}{x - 1}

簡素化

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 5 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{4}{x - 1} : \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

定数を解答に追加する

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C