integral of (4x+2)/((x+1)(x^2+1))

解

\int \frac{4 x + 2}{( x + 1 ) ( x ^{2} + 1 )} d x

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

解答ステップ

\int \frac{4 x + 2}{( x + 1 ) ( x ^{2} + 1 )} d x

拡張

\frac{4 x + 2}{( x + 1 ) ( x ^{2} + 1 )} : \frac{4 x}{( x + 1 ) ( x ^{2} + 1 )} + \frac{2}{( 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{4 x}{( x + 1 ) ( x ^{2} + 1 )} d x + \int \frac{2}{( x + 1 ) ( x ^{2} + 1 )} d x

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

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

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

簡素化

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

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

定数を解答に追加する

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