integral of (3x+3)/(x^2+7x+12)

解

\int \frac{3 x + 3}{x ^{2} + 7 x + 12} d x

6 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{7}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣)) + 3 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣) + C

解答ステップ

\int \frac{3 x + 3}{x ^{2} + 7 x + 12} d x

拡張

\frac{3 x + 3}{x ^{2} + 7 x + 12} : \frac{3 x}{x ^{2} + 7 x + 12} + \frac{3}{x ^{2} + 7 x + 12}

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{3 x}{x ^{2} + 7 x + 12} d x + \int \frac{3}{x ^{2} + 7 x + 12} d x

\int \frac{3 x}{x ^{2} + 7 x + 12} d x = 6 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{7}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣))

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

= 6 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{7}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣)) + 3 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣)

定数を解答に追加する

= 6 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{7}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣)) + 3 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣) + C