integral of (3x^4+4)/(x^2(x^2+1)^3)

解

\int \frac{3 x ^{4} + 4}{x ^{2} ( x ^{2} + 1 ) ^{3}} d x

- \frac{x}{2 ( x ^{2} + 1 )} - \frac{57}{8} arctan (x) - \frac{21}{8} \cdot \frac{x}{x ^{2} + 1} - \frac{4}{x} - \frac{7 x}{4 ( x ^{2} + 1 ) ^{2}} + C

解答ステップ

\int \frac{3 x ^{4} + 4}{x ^{2} ( x ^{2} + 1 ) ^{3}} d x

以下の部分分数を得る:

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

= \int \frac{4}{x ^{2}} - \frac{4}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{2}} - \frac{7}{( x ^{2} + 1 ) ^{3}} 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{4}{x ^{2}} d x - \int \frac{4}{x ^{2} + 1} d x - \int \frac{1}{( x ^{2} + 1 ) ^{2}} d x - \int \frac{7}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

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

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

\int \frac{7}{( x ^{2} + 1 ) ^{3}} d x = 7 (\frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}})

= - \frac{4}{x} - 4 arctan (x) - \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) - 7 (\frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}})

簡素化

- \frac{4}{x} - 4 arctan (x) - \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) - 7 (\frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}}) : - \frac{x}{2 ( x ^{2} + 1 )} - \frac{57}{8} arctan (x) - \frac{21}{8} \cdot \frac{x}{x ^{2} + 1} - \frac{4}{x} - \frac{7 x}{4 ( x ^{2} + 1 ) ^{2}}

= - \frac{x}{2 ( x ^{2} + 1 )} - \frac{57}{8} arctan (x) - \frac{21}{8} \cdot \frac{x}{x ^{2} + 1} - \frac{4}{x} - \frac{7 x}{4 ( x ^{2} + 1 ) ^{2}}

定数を解答に追加する

= - \frac{x}{2 ( x ^{2} + 1 )} - \frac{57}{8} arctan (x) - \frac{21}{8} \cdot \frac{x}{x ^{2} + 1} - \frac{4}{x} - \frac{7 x}{4 ( x ^{2} + 1 ) ^{2}} + C