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

解

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

\frac{3}{2} ln x^{2} + 1 + arctan (x) - ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

ln x^{2} + 1 + arctan (x) - (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) : \frac{3}{2} ln x^{2} + 1 + arctan (x) - ln ∣ x ∣

= \frac{3}{2} ln x^{2} + 1 + arctan (x) - ln ∣ x ∣

定数を解答に追加する

= \frac{3}{2} ln x^{2} + 1 + arctan (x) - ln ∣ x ∣ + C