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

解

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

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

解答ステップ

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

拡張

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

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

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

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

= - \frac{1}{x} - arctan (x) + 3 arctan (x)

簡素化

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

定数を解答に追加する

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