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

解

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

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

解答ステップ

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

簡素化

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

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

拡張

\frac{x + 1}{x ( 2 x + 3 )} : \frac{x}{x ( 2 x + 3 )} + \frac{1}{x ( 2 x + 3 )}

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

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

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

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

定数を解答に追加する

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