integral of (x^2+x+1)/((x^4+2x^2+1)^2)

解答

\int \frac{x ^{2} + x + 1}{( x ^{4} + 2 x ^{2} + 1 ) ^{2}} d x

\frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{6 ( x ^{2} + 1 ) ^{3}} + C

求解步骤

\int \frac{x ^{2} + x + 1}{( x ^{4} + 2 x ^{2} + 1 ) ^{2}} d x

将

\frac{x ^{2} + x + 1}{( x ^{4} + 2 x ^{2} + 1 ) ^{2}}

用部份分式展开:

\frac{1}{( x ^{2} + 1 ) ^{3}} + \frac{x}{( x ^{2} + 1 ) ^{4}}

= \int \frac{1}{( x ^{2} + 1 ) ^{3}} + \frac{x}{( x ^{2} + 1 ) ^{4}} 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} + 1 ) ^{3}} d x + \int \frac{x}{( x ^{2} + 1 ) ^{4}} d x

\int \frac{1}{( x ^{2} + 1 ) ^{3}} d x = \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}}

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

= \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{6 ( x ^{2} + 1 ) ^{3}}

解答补常数

= \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{6 ( x ^{2} + 1 ) ^{3}} + C

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

d er i v a t i v e f (x) = 2 x^{4} + 3 x + 16

\frac{d}{d t} (\frac{6}{5} - \frac{6}{5} e^{- 20 t})

a re a y = x^{3} - 6 x^{2} + 9 x, y = x, x = 0, x = 2

\int_{0}^{\infty} 4 x e^{- 2 x} d x