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

解答

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

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

求解步骤

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

乘开

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

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

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

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

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

化简

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

解答补常数

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

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