integral of (x^2-29x+5)/((x-4)^2(x^2+3))

解

\int \frac{x ^{2} - 29 x + 5}{( x - 4 ) ^{2} ( x ^{2} + 3 )} d x

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

解答ステップ

\int \frac{x ^{2} - 29 x + 5}{( x - 4 ) ^{2} ( x ^{2} + 3 )} d x

以下の部分分数を得る:

\frac{x ^{2} - 29 x + 5}{( x - 4 ) ^{2} ( x ^{2} + 3 )} : \frac{1}{x - 4} - \frac{5}{( x - 4 ) ^{2}} + \frac{- x + 2}{x ^{2} + 3}

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

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

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

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

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

簡素化

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

定数を解答に追加する

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