integral of (8x^2+x+9)/((x+1)(x^2+1))

解

\int \frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} d x

8 ln ∣ x + 1 ∣ + arctan (x) + C

解答ステップ

\int \frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} d x

拡張

\frac{8 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 1 )} : \frac{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} + \frac{x}{( x + 1 ) ( x ^{2} + 1 )} + \frac{9}{( x + 1 ) ( 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{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x + \int \frac{x}{( x + 1 ) ( x ^{2} + 1 )} d x + \int \frac{9}{( x + 1 ) ( x ^{2} + 1 )} d x

\int \frac{8 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x = 8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

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

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

= 8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 9 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

簡素化

8 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 9 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) : 8 ln ∣ x + 1 ∣ + arctan (x)

= 8 ln ∣ x + 1 ∣ + arctan (x)

定数を解答に追加する

= 8 ln ∣ x + 1 ∣ + arctan (x) + C