integral of (11x+8)/(10x^2+5x)

解

\int \frac{11 x + 8}{10 x ^{2} + 5 x} d x

\frac{11}{10} ln ∣ 2 x + 1 ∣ - \frac{8}{5} (ln ∣ 4 x + 2 ∣ - ln (4 ∣ x ∣)) + C

解答ステップ

\int \frac{11 x + 8}{10 x ^{2} + 5 x} d x

拡張

\frac{11 x + 8}{10 x ^{2} + 5 x} : \frac{11 x}{10 x ^{2} + 5 x} + \frac{8}{10 x ^{2} + 5 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{11 x}{10 x ^{2} + 5 x} d x + \int \frac{8}{10 x ^{2} + 5 x} d x

\int \frac{11 x}{10 x ^{2} + 5 x} d x = \frac{11}{10} ln ∣ 2 x + 1 ∣

\int \frac{8}{10 x ^{2} + 5 x} d x = - \frac{8}{5} (ln ∣ 4 x + 2 ∣ - ln (4 ∣ x ∣))

= \frac{11}{10} ln ∣ 2 x + 1 ∣ - \frac{8}{5} (ln ∣ 4 x + 2 ∣ - ln (4 ∣ x ∣))

定数を解答に追加する

= \frac{11}{10} ln ∣ 2 x + 1 ∣ - \frac{8}{5} (ln ∣ 4 x + 2 ∣ - ln (4 ∣ x ∣)) + C