integral of (9x^2-8x+8)/(x(x^2+4))

解

\int \frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} d x

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

解答ステップ

\int \frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} d x

拡張

\frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} : \frac{9 x ^{2}}{x ( x ^{2} + 4 )} - \frac{8 x}{x ( x ^{2} + 4 )} + \frac{8}{x ( x ^{2} + 4 )}

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

\int \frac{9 x ^{2}}{x ( x ^{2} + 4 )} d x = \frac{9}{2} ln x^{2} + 4

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

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

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ - ln x^{2} + 4

類似した元を足す:

\frac{9}{2} ln x^{2} + 4 - ln x^{2} + 4 = \frac{7}{2} ln x^{2} + 4

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

定数を解答に追加する

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