integral of (x^2+3)/(x^6(x^2+1))

解答

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

- \frac{2}{x} + \frac{2}{3 x ^{3}} - \frac{3}{5 x ^{5}} - 2 arctan (x) + C

求解步骤

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

将

\frac{x ^{2} + 3}{x ^{6} ( x ^{2} + 1 )}

用部份分式展开:

\frac{2}{x ^{2}} - \frac{2}{x ^{4}} + \frac{3}{x ^{6}} - \frac{2}{x ^{2} + 1}

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

\int \frac{2}{x ^{2}} d x = - \frac{2}{x}

\int \frac{2}{x ^{4}} d x = - \frac{2}{3 x ^{3}}

\int \frac{3}{x ^{6}} d x = - \frac{3}{5 x ^{5}}

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

= - \frac{2}{x} - (- \frac{2}{3 x ^{3}}) - \frac{3}{5 x ^{5}} - 2 arctan (x)

化简

= - \frac{2}{x} + \frac{2}{3 x ^{3}} - \frac{3}{5 x ^{5}} - 2 arctan (x)

解答补常数

= - \frac{2}{x} + \frac{2}{3 x ^{3}} - \frac{3}{5 x ^{5}} - 2 arctan (x) + C

\int \frac{( x + 1 )}{3 x ^{2} + 6 x + 5} d x

\int 7 - 2 x^{2} d x

\int 4 sin (x) cos^{3} (x) d x

\int \frac{1}{64} e^{4 x} d x

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