
第一节不定积分的换元积分法
第二节 不定积分的换元积分法

第一类换元法(微分法)设定理2.1[ f(x)dx = F(x)+C,则 [ f(0(x)g'(x)dx=[ f(o(x)dp(x)= F(Φ(x)+ C
第一类换元法(凑微分法) ( ) ( ) , ( ( )) ( ) ( ( )) ( ) ( ( )) . f x dx F x C f x x dx f x d x F x C = + = = + 定理2.1 设 则

例1 求[(3x + 2)’dx.[(3x + 2)'dx =3x +2rdt解3.1+C-B1-(3x + 2)° + C.18
5 1 (3 2 x dx + ) . 例 求 5 5 6 6 3 2 1 (3 2) 3 1 18 1 (3 2) . 18 t x x dx t dt t C x C = + + = + = + + 解

例2 求 xe" dx.解『xe"'dxe'd-12-12+C+ C
2 2 . x xe dx 例 求 2 2 2 1 2 1 2 1 . 2 x t t x t x xe dx e dt e C e C = = + = + 解

1例3 求[-dx.x(1+ 2lnx)1解d(1+ 2ln x)dx =271+2lnxx(1+ 2ln x)- 1ml+ 1n + c
1 . (1 2 ) 3 ln dx x x + 例 求 1 1 1 (1 2ln ) (1 2ln ) 2 1 2ln 1 ln 1 2ln . 2 dx d x x x x x C = + + + = + + 解

dx4 求[例45+ x0dxdx解0Yarctan- + C.a
2 2 4 . dx a x + 例 求 2 2 2 2 2 1 1 1 1 arctan . dx dx a x x a a x d a a x a x C a a = + + = + = + 解

1例5 求[dx.x2 +2x+211解dxdx =2x2 +2x +21+(x+1)1= 1+(+1yd(++1)= arctan(x + 1)+ C
2 1 . 2 2 5 dx x x + + 例 求 2 2 2 1 1 2 2 1 ( 1) 1 ( 1) 1 ( 1) arctan( 1) . dx dx x x x d x x x C = + + + + = + + + = + + 解

1例6 求dx (a > 0).- x解dxdx =1-()1-= arcsin - + C.a
2 2 1 6 dx a( 0). a x − 例 求 2 2 2 2 1 1 1 1 1 arcsin . dx dx a x x a a x d a x a x C a = − − = − = + 解

1例7 求[dx.V1-2x -x1解dx =d(x + 1)V1-2x-x[2 -(x +1)x+1+ C.= arcsinV2
2 7 1 . 1 2 dx − − x x 例 求 2 2 1 1 ( 1) 1 2 2 ( 1) 1 arcsin . 2 dx d x x x x x C = + − − − + + = + 解

dx例8 求[2 - x2dx解2aJ(dx一++dx+-dx2aa+xa-x--2 --- Ina-x|+C0a+x2aa+x+ C.In2aa-x
2 2 8 . dx a x − 例 求 2 2 1 1 1 2 1 1 1 1 2 2 1 1 ln ln 2 2 1 ln . 2 dx dx a x a a x a x dx dx a a x a a x a x a x C a a a x C a a x = + − + − = + + − = + − − + + = + − 解