integral of (x^2)/(x^4+10x^2+9)

解答

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x

- \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) + C

求解步骤

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x

分解

\frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9}

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

将

\frac{x ^{2}}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

用部份分式展开:

- \frac{1}{8 ( x ^{2} + 1 )} + \frac{9}{8 ( x ^{2} + 9 )}

= \int - \frac{1}{8 ( x ^{2} + 1 )} + \frac{9}{8 ( x ^{2} + 9 )} 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}{8 ( x ^{2} + 1 )} d x + \int \frac{9}{8 ( x ^{2} + 9 )} d x

\int \frac{1}{8 ( x ^{2} + 1 )} d x = \frac{1}{8} arctan (x)

\int \frac{9}{8 ( x ^{2} + 9 )} d x = \frac{3}{8} arctan (\frac{x}{3})

= - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3})

解答补常数

= - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) + C

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d er i v a t i v e f (x) = \frac{1}{7 - x}

\frac{d}{d x} (h (x))

d er i v a t i v e f (x) = \frac{9}{x}

\frac{d y}{d x} = \frac{1}{6} y cos^{2} (y)