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

解

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

\frac{4 x ^{3}}{3} - 2 x^{2} + 23 x - 94 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + 138 (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣) + C

解答ステップ

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

長除法

\frac{4 x ^{4} - 5 x ^{2}}{( x + 3 ) ( x - 2 )} : 4 x^{2} - 4 x + 23 + \frac{- 47 x + 138}{x ^{2} + x - 6}

= \int 4 x^{2} - 4 x + 23 + \frac{- 47 x + 138}{x ^{2} + x - 6} 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 4 x^{2} d x - \int 4 x d x + \int 23 d x + \int \frac{- 47 x + 138}{x ^{2} + x - 6} d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 23 d x = 23 x

\int \frac{- 47 x + 138}{x ^{2} + x - 6} d x = - 94 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + 138 (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣)

= \frac{4 x ^{3}}{3} - 2 x^{2} + 23 x - 94 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + 138 (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣)

定数を解答に追加する

= \frac{4 x ^{3}}{3} - 2 x^{2} + 23 x - 94 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + 138 (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣) + C