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

解

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

\frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49} + C

解答ステップ

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

拡張

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

\int \frac{x ^{2}}{3 x ^{2} + 4 x ^{2} - 4 x} d x = \frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣)

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

= \frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣) - ln ∣ x ∣ + ln ∣ 7 x - 4 ∣

簡素化

\frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣) - ln ∣ x ∣ + ln ∣ 7 x - 4 ∣ : \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49}

= \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49}

定数を解答に追加する

= \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49} + C