例2求 3+2x 解 (3+2x), 3+2x23+2x 3+2x2J3+2x (3+2x)dx d(3+2x) 23+2x du==lnu+C==In 3+2x+C 2 般地f(ax+b)d=Uf(a)dl lu=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 (3 2 ) 3 2 1 2 1 d x x + + =
例3求 C。 x(1+2nx) 解 x(1+2nx) d(nx) 1+2Inx d(1+2Inx) 2 1+2Inx u=1+2nx 2 ∫hm=lma+C=,h+2hx+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 = 1+ 2ln x = 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、4=∫A (1+x 1+x) 3 (1+x)2(1+x)3 Ja(1+x) 十 C,+ +c 1+x 2(1+x) 1+x2(1+吵+C
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「,,b a+x 解Ja+x=a2 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8求 解 2 2 a arcsin -+C
例8 求 − 2 2 a x dx 解： − 2 2 a x dx 2 1 ( ) 1 a x dx a − = − = 2 1 ( ) a x a x d c a x = arcsin +