integral of (x^2-x+3)/(x^4+6x^2+9)

解答

\int \frac{x ^{2} - x + 3}{x ^{4} + 6 x ^{2} + 9} d x

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

求解步骤

\int \frac{x ^{2} - x + 3}{x ^{4} + 6 x ^{2} + 9} d x

将

\frac{x ^{2} - x + 3}{x ^{4} + 6 x ^{2} + 9}

用部份分式展开:

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

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

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

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

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

化简

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

解答补常数

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

\int 5 - 3 x^{3} + 2 x^{2} d x

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