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

解答

\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

求解步骤

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

乘开

\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 )}

使用积分加法定则:

\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)

化简

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 ∣

解答补常数

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

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

x^{2} + 2 y^{2} - 2 x - 8 y + 6 = 0 (0.3)

\int \frac{36 x ^{2} - 49}{x ^{3}} d x

x \to \frac{π}{2} - lim (sec (x))

t an g e n t - 9 x^{\frac{1}{3}} + 5, at x = 27