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

解答

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

\frac{1}{x} + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3})) + C

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

\int \frac{- x + 1}{3 ( x ^{2} + x + 1 )} d x = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

= - (- \frac{1}{x}) + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

化简

- (- \frac{1}{x}) + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3})) : \frac{1}{x} + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

= \frac{1}{x} + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

解答补常数

= \frac{1}{x} + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3})) + C

\int \frac{x}{x ^{2} - 2 x + 3} d x

\int \frac{1}{sin ( x ) - tan ( x )} d x

\int 3 x^{2} (x^{3} + 4)^{5} d x

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

\int x^{2} (4 - x^{3})^{\frac{2}{3}} d x