展开 ((1+y^2)(-2x-arctan(y)))/(x-y^2-1)

解答

展开

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

\frac{- 2 x - arctan ( y ) - 2 y ^{2} x - y ^{2} arctan ( y )}{x - y ^{2} - 1}

求解步骤

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

乘开

(1 + y^{2}) (- 2 x - arctan (y)) : - 2 x - arctan (y) - 2 y^{2} x - y^{2} arctan (y)

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

e x p an d (cos (x) - sin (x)) \cdot cos (x)

s im pl i f y - i x (2 - i) (5 x + 6 i)

e x p an d - \frac{1}{s + 1} - 2 s - 2 + (\frac{1}{s + 1})

e x p an d 3 (5 + 13)

s im pl i f y (8 x - 3) (6 x + 2)