integral of (-4x^3)/(2x^3+x^2-2x-1)

解

\int \frac{- 4 x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

- 4 (\frac{x ^{4}}{2 x ^{3} + x ^{2} - 2 x - 1} + \frac{1}{12} ln ∣ 2 x + 1 ∣ + \frac{1}{12 ( 2 x + 1 )} - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6 ( x - 1 )}) + C

解答ステップ

\int \frac{- 4 x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 4 \cdot \int \frac{x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

部分積分を適用する

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

\int \frac{x ( x ^{4} - 4 x ^{3} - 3 x ^{2} )}{( 2 x ^{3} + x ^{2} - 2 x - 1 ) ^{2}} d x = - \frac{1}{12} ln ∣ 2 x + 1 ∣ - \frac{1}{12 ( 2 x + 1 )} + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{6 ( x - 1 )}

= - 4 (\frac{x ^{4}}{2 x ^{3} + x ^{2} - 2 x - 1} - (- \frac{1}{12} ln ∣ 2 x + 1 ∣ - \frac{1}{12 ( 2 x + 1 )} + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{6 ( x - 1 )}))

簡素化

= - 4 (\frac{x ^{4}}{2 x ^{3} + x ^{2} - 2 x - 1} + \frac{1}{12} ln ∣ 2 x + 1 ∣ + \frac{1}{12 ( 2 x + 1 )} - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6 ( x - 1 )})

定数を解答に追加する

= - 4 (\frac{x ^{4}}{2 x ^{3} + x ^{2} - 2 x - 1} + \frac{1}{12} ln ∣ 2 x + 1 ∣ + \frac{1}{12 ( 2 x + 1 )} - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6 ( x - 1 )}) + C