integral of (6x^2-3x+1)/((4x+1)(x^2+1))

解

\int \frac{6 x ^{2} - 3 x + 1}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

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

解答ステップ

\int \frac{6 x ^{2} - 3 x + 1}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

以下の部分分数を得る:

\frac{6 x ^{2} - 3 x + 1}{( 4 x + 1 ) ( x ^{2} + 1 )} : \frac{2}{4 x + 1} + \frac{x - 1}{x ^{2} + 1}

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

\int \frac{2}{4 x + 1} d x = \frac{1}{2} ln ∣ 4 x + 1 ∣

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

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

定数を解答に追加する

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