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

解

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

ln ∣ x ∣ - ln 1 + x^{2} + C

解答ステップ

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

拡張

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

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

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

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

類似した元を足す:

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

= ln ∣ x ∣ - ln 1 + x^{2}

定数を解答に追加する

= ln ∣ x ∣ - ln 1 + x^{2} + C