integral of 8x^2+6+9/(x^2+1)

解

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 1} d x

\frac{8 x ^{3}}{3} + 6 x + 9 arctan (x) + C

解答ステップ

\int 8 x^{2} + 6 + \frac{9}{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

= \int 8 x^{2} d x + \int 6 d x + \int \frac{9}{x ^{2} + 1} d x

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 6 d x = 6 x

\int \frac{9}{x ^{2} + 1} d x = 9 arctan (x)

= \frac{8 x ^{3}}{3} + 6 x + 9 arctan (x)

定数を解答に追加する

= \frac{8 x ^{3}}{3} + 6 x + 9 arctan (x) + C