integral of (8x^3-16x^2+5)/(x^2-2x)

解

\int \frac{8 x ^{3} - 16 x ^{2} + 5}{x ^{2} - 2 x} d x

4 x^{2} + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x ∣ - 16 + C

解答ステップ

\int \frac{8 x ^{3} - 16 x ^{2} + 5}{x ^{2} - 2 x} d x

拡張

\frac{8 x ^{3} - 16 x ^{2} + 5}{x ^{2} - 2 x} : \frac{8 x ^{3}}{x ^{2} - 2 x} - \frac{16 x ^{2}}{x ^{2} - 2 x} + \frac{5}{x ^{2} - 2 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{8 x ^{3}}{x ^{2} - 2 x} d x - \int \frac{16 x ^{2}}{x ^{2} - 2 x} d x + \int \frac{5}{x ^{2} - 2 x} d x

\int \frac{8 x ^{3}}{x ^{2} - 2 x} d x = 8 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣)

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

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

= 8 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 16 (x - 2 + 2 ln ∣ x - 2 ∣) - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

簡素化

8 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 16 (x - 2 + 2 ln ∣ x - 2 ∣) - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) : 4 x^{2} + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x ∣ - 16

= 4 x^{2} + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x ∣ - 16

定数を解答に追加する

= 4 x^{2} + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x ∣ - 16 + C