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

解

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

- \frac{1}{2 2} arctan (2 x) + \frac{1}{2} x + 2 arctan (2 x) + C

解答ステップ

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

拡張

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

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

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

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

定数を解答に追加する

= - \frac{1}{2 2} arctan (2 x) + \frac{1}{2} x + 2 arctan (2 x) + C