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

解答

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

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

求解步骤

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

乘开

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

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

\int \frac{x arctan ( x )}{( 1 + x ^{2} ) ^{2}} d x = \frac{1}{2} (- \frac{1}{2} arctan (x) cos (2 arctan (x)) + \frac{1}{4} sin (2 arctan (x)))

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

解答补常数

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

\int (sin (3 x) cos (3 x)) d x

\int \frac{x ^{2}}{ln ^{2} ( x )} d x

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

\int x^{2} (x^{2} - 3) d x

\int \frac{2 + x}{x} d x