integral of ((x^2+x))/((x-3)(x^2+1))

解

\int \frac{( x ^{2} + x )}{( x - 3 ) ( x ^{2} + 1 )} d x

\frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x) + C

解答ステップ

\int \frac{x ^{2} + x}{( x - 3 ) ( x ^{2} + 1 )} d x

拡張

\frac{x ^{2} + x}{( x - 3 ) ( x ^{2} + 1 )} : \frac{x ^{2}}{( x - 3 ) ( x ^{2} + 1 )} + \frac{x}{( x - 3 ) ( x ^{2} + 1 )}

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{x ^{2}}{( x - 3 ) ( x ^{2} + 1 )} d x + \int \frac{x}{( x - 3 ) ( x ^{2} + 1 )} d x

\int \frac{x ^{2}}{( x - 3 ) ( x ^{2} + 1 )} d x = \frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x))

\int \frac{x}{( x - 3 ) ( x ^{2} + 1 )} d x = \frac{3}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{3}{2} ln x^{2} + 1 + arctan (x))

= \frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x)) + \frac{3}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{3}{2} ln x^{2} + 1 + arctan (x))

簡素化

\frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x)) + \frac{3}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{3}{2} ln x^{2} + 1 + arctan (x)) : \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x)

= \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x)

定数を解答に追加する

= \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x) + C