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

解

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

x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C

解答ステップ

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

長除法

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

= \int 1 + \frac{x - 1}{4 x ^{2} - 4 x + 3} 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 1 d x + \int \frac{x - 1}{4 x ^{2} - 4 x + 3} d x

\int 1 d x = x

\int \frac{x - 1}{4 x ^{2} - 4 x + 3} d x = \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

= x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

定数を解答に追加する

= x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C