integral of (6x^4+24x^2+4)/(x^3+4x)

解

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12 + C

解答ステップ

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

拡張

\frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} : \frac{6 x ^{4}}{x ^{3} + 4 x} + \frac{24 x ^{2}}{x ^{3} + 4 x} + \frac{4}{x ^{3} + 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{6 x ^{4}}{x ^{3} + 4 x} d x + \int \frac{24 x ^{2}}{x ^{3} + 4 x} d x + \int \frac{4}{x ^{3} + 4 x} d x

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

\int \frac{24 x ^{2}}{x ^{3} + 4 x} d x = 12 ln x^{2} + 4

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

= 3 (x^{2} + 4 - 4 ln x^{2} + 4) + 12 ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

簡素化

3 (x^{2} + 4 - 4 ln x^{2} + 4) + 12 ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12

= 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12

定数を解答に追加する

= 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12 + C