例2求 3+2x 解 (3+2x), 3+2x 2 3+2x = 3+2x2J3+2x (3+2x) du =Inu+C=In(3+ 2x)+C 2 2 般地∫f(ax+bhl ∫(u)ldn u=ax+b
例2 求 . 3 2 1 dx x + 解 (3 2 ) , 3 2 1 2 1 3 2 1 + + = + x x x dx x 3 + 2 1 x dx x (3 2 ) 3 2 1 2 1 + + = du u = 1 2 1 = lnu + C 2 1 ln(3 2 ) . 2 1 = + x + C f (ax + b)dx = u du u=ax+b f a [ ( ) ] 1 一般地
例3求 x x(1+2In x 解 x(1+2nx) d=∫,dmx) +2In d(1+2nx) 1+2Inx u=1+2nx du =Inu+C=In(1+2Inx)+C
例3 求 . (1 2ln ) 1 dx x x + 解 dx x x (1+ 2ln ) 1 (ln ) 1 2ln 1 d x x + = (1 2ln ) 1 2ln 1 2 1 d x x + + = u x = +1 2ln = du u 1 2 1 = lnu + C 2 1 ln(1 2ln ) . 2 1 = + x + C
例4求 (1+x) x+1-1 解 (x= (1+x)3 3 1+x) (1+x)2(1+x)3 Id(1+x) +C, +c 1+x 2(1+x) 2 + 1+x2(1+x)
例4 求 . (1 ) 3 dx x x + 解 dx 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「,,d 2 a+x 解 a+x arctan -+C 1+
例5 求 . 1 2 2 dx a x + 解 dx a x + 2 2 1 dx 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 = +
例6 2-8x+25 解 x2-8x+25 x-4)2+9 x-4 32/x-4 +1 +1 3 3 =arctan +c 3
例6 求 . 8 25 1 2 dx x x − + 解 dx x x − 8 + 25 1 2 dx x − + = ( 4) 9 1 2 dx 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 + − =