integral of (7x^2)/((x+1)(x^2+1))

解

\int \frac{7 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

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

解答ステップ

\int \frac{7 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int \frac{x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

以下の部分分数を得る:

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

= 7 \cdot \int \frac{1}{2 ( x + 1 )} + \frac{x - 1}{2 ( x ^{2} + 1 )} d 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

= 7 (\int \frac{1}{2 ( x + 1 )} d x + \int \frac{x - 1}{2 ( x ^{2} + 1 )} d x)

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

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

= 7 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

定数を解答に追加する

= 7 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) + C