(11) d(=) 2+x22 1 1+()2 (12) -d(-)=arcsin +C 2 2.计算下列积分 arctan 2x (3)「xedx (4)e"sin 4xdx, (5)xsin 100xdx (6)arctan 2xdx 解:(1)「h2xdx=xh2x-Jxd(h2x) xh2x-∫x.dx xIn 2x-x+C (2)arctan 2xdx=x arctan 2x- xd(arctan x) = x arctan2x-∫x 1+(2x) x arctan 2x ∫,x 1+4x =arctan 2x- ,d+4x2) 4J1+4x =arctan 2x--In(1+4x)+C (3) dy 4 (4)Je sin 4xdx= sin 4xd(=)==e sin 4x-= d( sin 4x) =ex sin 4x==J cos 4.e3 cos 4x ∫cdcs4(11) x C x x x x x x = + + = + = + 2 2 arctan 2 2 ) 2 d( ) 2 1 ( 1 2 1 ) 2 1 ( d 2 1 2 d 2 2 2 . (12) 2 4 - d x x = 2 ) 2 2 1-( d x x = ) 2 d( ) 2 1-( 1 2 x x = C x + 2 arcsin . 2. 计算下列积分: (1) ln 2xdx , (2) arctan 2xdx , (3) x x x e d 4 , (4) x x x e sin 4 d 5 , (5) x sin 100xdx , (6) x arctan 2xdx . 解:(1) ln 2xdx = x ln 2x − xd(ln 2x) = x x x x x d 2 2 ln 2 − = xln 2x − x +C. (2) arctan 2xdx = x arctan 2x − xd(arctan2 x) = x x x x x d 1 (2 ) 2 arctan 2 2 + − = + − 2 2 1 4 d( ) arctan 2 x x x x = d(1 4 ) 1 4 1 4 1 arctan 2 2 2 x x x x + + − = x x − ln(1+ 4x ) + C 4 1 arctan 2 2 . (3) x x x x x x x x x e d 4 1 e 4 1 de 4 1 e d 4 4 4 4 = = − = x C x x − + 4 4 e 16 1 e 4 1 . (4) 5 5 5 5 e 1 e e sin 4 d sin 4 d( ) e sin 4 d(sin 4 ) 5 5 5 x x x x = = − x x x x x = x x x x x e cos 4 d 5 4 e sin 4 5 1 5 5 − = 5 e cos 4 d 5 4 e sin 4 5 1 5 5 x x x − x = − − d(cos 4 ) 5 e cos 4 5 e 5 4 e sin 4 5 1 5 5 5 x x x x x x