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(7) (8)(2x+3)dx,(9) arcsin√1-x (10) dx,(11) (12) (1+x2) 解:(1) d(sin x) C 6 2)∫cos3xdx= )d(sin x) +c (3)J(x+ x-2cos√x+C ∫xe'dx= Jed(x)=ex+c (5 ∫(1-x2)2d(1 C I d x2+c xdr=[ n2 ∫h2xd(h2x)=h2x+ d(arcsin x)=h arcsin x +C arcsin x d(arctan x)=In arctan x +C (+x)arctan x arctan x(4) xe x x d 2  , (5)  − 2 1 d x x x , (6)  − 4 1 d x x x , (7)  x x x d ln 2 , (8) (2x 3) dx 2  + , (9)  −  dx x x 2 1 1 arcsin 1 , (10)  + x x x d (1 ) arctan 1 2 , (11)  + 2 2 d x x , (12)  − 2 4 d x x . 解:(1) C x  x x = + 6 sin sin d(sin ) 6 5 . (2) cos x dx (1 sin x)cos x dx 3 2  =  − = (1 sin )d(sin ) 2  − x x = d(sin ) sin d(sin ) 2  x −  x x = C x x − + 3 sin sin 3 . (3) x x x x x x x x )d d 2 sin d sin ( + =  +  = x C x − 2cos + 2 2 . (4) x x x C x x x  =  = + 2 2 2 e 2 1 e d( ) 2 1 e d 2 . (5) x x x x C x x = −  − − = − − + − −  2 2 2 1 2 2 (1 ) d(1 ) 1 2 1 d 1 . (6) x C x x x x x = + − = −   2 2 2 2 4 arcsin 2 1 1 ( ) d( ) 2 1 1 d . (7) x x x x C x x x x x = =  = +   ln 2 2 1 d(2 ) ln 2 d(ln 2 ) 2 ln 2 d ln 2 2 . (8)  x + x =  x + x + = x + + C 2 2 3 (2 3) 6 1 (2 3) d(2 3) 2 1 (2 3) d . (9) x x C x x x x = = + −    d(arcsin ) ln | arcsin | arcsin 1 d 1 1 arcsin 1 2 . (10) x x C x x x x = = + +   d(arctan ) ln | arctan | arctan 1 d (1 ) arctan 1 2
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