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

解

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

\frac{2 x ^{3}}{3} - x + 4 arctan (x) + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

= \frac{2 x ^{3}}{3} - x + 4 arctan (x) + C