例2求 ax 3+2x 解 (3+2x), 3+2x23+2x dx= 3+2x 23+2(3+2x)yx du =Inu+C=In(3+2x)+C 2u 2 2 u=ax tb 王页下
例2 求 . 3 2 1 d x x + 解 (3 2 ) , 3 2 1 2 1 3 2 1 + + = + x x x d x x 3 + 2 1 x d x x (3 2 ) 3 2 1 2 1 + + = du u = 1 2 1 = ln u + C 2 1 ln(3 2 ) . 2 1 = + x + C f (a x + b)d x = u=ax +b f u d u a [ ( ) ] 1 一般地
例3求 dx x(1+2In x) 解 dx x(1+2In x) d(n x) 1+2Inx d(1+ 2In x) 21+2In x u=1+2Inx 1。1 a-du =Inu+C= In(1+2Inx+C L 2 2 上页
例3 求 . (1 2 ln ) 1 d x x x + 解 d x x x (1 + 2 ln ) 1 (ln ) 1 2ln 1 d x x + = (1 2ln ) 1 2ln 1 2 1 d x x + + = u = 1+ 2ln x = du u 1 2 1 = ln u + C 2 1 ln(1 2ln ) . 2 1 = + x + C
例4求 dx (1+x) +1-1 解 x dx (1+x)3 - dx 3 (1+x) ∫ 1 1+x)2(1+1)31(1+x) 1 +C, 十 e +c 1+x 2(1+x)22 1+C 1+x2(1+x) 上页
例4 求 . (1 ) 3 d x x x + 解 d x x x + 3 (1 ) dx x x + + − = 3 (1 ) 1 1 ] (1 ) (1 ) 1 (1 ) 1 [ 2 3 d x x x + + − + = 1 2 2 2(1 ) 1 1 1 C x C x + + + + + = − . 2(1 ) 1 1 1 2 C x x + + + + = −
例 5 求 十 解 d 十 =arctan+ C
例5 求 . 1 2 2 d x a x + 解 d x a x + 2 2 1 d x a a x + = 2 2 2 1 1 1 + = a x d a a x 2 1 1 1 arctan . 1 C a x a = +
M例6求2 x2-8x+25 解∫ dx x2-8x+25 1dx (x-4)2+9 x-4 3 d x 3 2 工工工 +1 十 3 3 x-4 =arctan +C 3 3 上页
例6 求 . 8 2 5 1 2 d x x x − + 解 d x x x − 8 + 2 5 1 2 dx x − + = ( 4) 9 1 2 d x x + − = 1 3 4 1 3 1 2 2 − + − = 3 4 1 3 4 1 3 1 2 x d x . 3 4 arctan 3 1 C x + − =