integral of (12x^2-4x)/((x^2+4)(x+2))

解

\int \frac{12 x ^{2} - 4 x}{( x ^{2} + 4 ) ( x + 2 )} d x

\frac{5}{2} ln \frac{x ^{2}}{4} + 1 - 7 arctan (\frac{x}{2}) + 7 ln ∣ x + 2 ∣ + C

解答ステップ

\int \frac{12 x ^{2} - 4 x}{( x ^{2} + 4 ) ( x + 2 )} d x

拡張

\frac{12 x ^{2} - 4 x}{( x ^{2} + 4 ) ( x + 2 )} : \frac{12 x ^{2}}{( x ^{2} + 4 ) ( x + 2 )} - \frac{4 x}{( x ^{2} + 4 ) ( x + 2 )}

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{12 x ^{2}}{( x ^{2} + 4 ) ( x + 2 )} d x - \int \frac{4 x}{( x ^{2} + 4 ) ( x + 2 )} d x

\int \frac{12 x ^{2}}{( x ^{2} + 4 ) ( x + 2 )} d x = 12 (\frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣)

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

= 12 (\frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣) - (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}) - ln ∣ x + 2 ∣)

簡素化

12 (\frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣) - (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}) - ln ∣ x + 2 ∣) : \frac{5}{2} ln \frac{x ^{2}}{4} + 1 - 7 arctan (\frac{x}{2}) + 7 ln ∣ x + 2 ∣

= \frac{5}{2} ln \frac{x ^{2}}{4} + 1 - 7 arctan (\frac{x}{2}) + 7 ln ∣ x + 2 ∣

定数を解答に追加する

= \frac{5}{2} ln \frac{x ^{2}}{4} + 1 - 7 arctan (\frac{x}{2}) + 7 ln ∣ x + 2 ∣ + C