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

解

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

- ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

= \int - \frac{1}{x} + \frac{5}{2 ( 3 x - 2 )} + \frac{1}{2 ( 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{1}{x} d x + \int \frac{5}{2 ( 3 x - 2 )} d x + \int \frac{1}{2 ( x + 2 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{2 ( 3 x - 2 )} d x = \frac{5}{6} ln ∣ 3 x - 2 ∣

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

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

定数を解答に追加する

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C