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

解

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

- \frac{7}{4} ln ∣ x ∣ - \frac{3}{2 x} + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{13}{12} ln ∣ x + 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{5 x - 3}{x ^{4} + x ^{3} - 2 x ^{2}} : - \frac{7}{4 x} + \frac{3}{2 x ^{2}} + \frac{2}{3 ( x - 1 )} + \frac{13}{12 ( x + 2 )}

= \int - \frac{7}{4 x} + \frac{3}{2 x ^{2}} + \frac{2}{3 ( x - 1 )} + \frac{13}{12 ( x + 2 )} 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{7}{4 x} d x + \int \frac{3}{2 x ^{2}} d x + \int \frac{2}{3 ( x - 1 )} d x + \int \frac{13}{12 ( x + 2 )} d x

\int \frac{7}{4 x} d x = \frac{7}{4} ln ∣ x ∣

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

\int \frac{2}{3 ( x - 1 )} d x = \frac{2}{3} ln ∣ x - 1 ∣

\int \frac{13}{12 ( x + 2 )} d x = \frac{13}{12} ln ∣ x + 2 ∣

= - \frac{7}{4} ln ∣ x ∣ - \frac{3}{2 x} + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{13}{12} ln ∣ x + 2 ∣

定数を解答に追加する

= - \frac{7}{4} ln ∣ x ∣ - \frac{3}{2 x} + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{13}{12} ln ∣ x + 2 ∣ + C