integral of (5x^4+7x^2+x+2)/(x(x^2+1)^2)

解

\int \frac{5 x ^{4} + 7 x ^{2} + x + 2}{x ( x ^{2} + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{5 x ^{4} + 7 x ^{2} + x + 2}{x ( x ^{2} + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{5 x ^{4} + 7 x ^{2} + x + 2}{x ( x ^{2} + 1 ) ^{2}} : \frac{2}{x} + \frac{3 x}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{2}}

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{3 x}{x ^{2} + 1} d x = \frac{3}{2} ln x^{2} + 1

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

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

定数を解答に追加する

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