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

解

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

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

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

簡素化

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

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

\frac{x ^{2}}{1 + x ^{2}} = - \frac{1}{1 + x ^{2}} + 1

= \frac{1}{2} \cdot \int - \frac{1}{1 + 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

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

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

\int 1 d x = x

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

定数を解答に追加する

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