integral of (7x^2-6x+8)/(x^3+4x)

解

\int \frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} d x

\frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ + C

解答ステップ

\int \frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} d x

拡張

\frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} : \frac{7 x ^{2}}{x ^{3} + 4 x} - \frac{6 x}{x ^{3} + 4 x} + \frac{8}{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{7 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{6 x}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

\int \frac{7 x ^{2}}{x ^{3} + 4 x} d x = \frac{7}{2} ln x^{2} + 4

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

\int \frac{8}{x ^{3} + 4 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 4

= \frac{7}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ - ln x^{2} + 4

類似した元を足す:

\frac{7}{2} ln x^{2} + 4 - ln x^{2} + 4 = \frac{5}{2} ln x^{2} + 4

= \frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣

定数を解答に追加する

= \frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ + C