integral of (5x^2+x-34)/((x+1)^2(x-4))

解

\int \frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} d x

3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣ + C

解答ステップ

\int \frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} d x

以下の部分分数を得る:

\frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} : \frac{3}{x + 1} + \frac{6}{( x + 1 ) ^{2}} + \frac{2}{x - 4}

= \int \frac{3}{x + 1} + \frac{6}{( x + 1 ) ^{2}} + \frac{2}{x - 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{3}{x + 1} d x + \int \frac{6}{( x + 1 ) ^{2}} d x + \int \frac{2}{x - 4} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{6}{( x + 1 ) ^{2}} d x = - \frac{6}{x + 1}

\int \frac{2}{x - 4} d x = 2 ln ∣ x - 4 ∣

= 3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣

定数を解答に追加する

= 3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣ + C