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

解答

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

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

求解步骤

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

乘开

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

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

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

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

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

解答补常数

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

\int x^{2} (x^{3} + 1)^{9} d x

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