integral of (5x+4)/((x+8)^2(x-1))

解

\int \frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} d x

- \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8} + C

解答ステップ

\int \frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} d x

拡張

\frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} : \frac{5 x}{( x + 8 ) ^{2} ( x - 1 )} + \frac{4}{( x + 8 ) ^{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{5 x}{( x + 8 ) ^{2} ( x - 1 )} d x + \int \frac{4}{( x + 8 ) ^{2} ( x - 1 )} d x

\int \frac{5 x}{( x + 8 ) ^{2} ( x - 1 )} d x = 5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

\int \frac{4}{( x + 8 ) ^{2} ( x - 1 )} d x = 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

= 5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) + 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

簡素化

5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) + 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) : - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8}

= - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8}

定数を解答に追加する

= - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8} + C