integral of ((4x+3))/(4x^3+8x^2+3x)

解

\int \frac{( 4 x + 3 )}{4 x ^{3} + 8 x ^{2} + 3 x} d x

ln ∣ x ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x + 3 ∣ + C

解答ステップ

\int \frac{4 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

以下の部分分数を得る:

\frac{4 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} : \frac{1}{x} - \frac{1}{2 x + 1} - \frac{1}{2 x + 3}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{2 x + 3} d x = \frac{1}{2} ln ∣ 2 x + 3 ∣

= ln ∣ x ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x + 3 ∣

定数を解答に追加する

= ln ∣ x ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x + 3 ∣ + C