高等数学A上册习题解答 至微分中值定理(6)

华中师范大学专用

µ

1

=

cosxdx1

2du1 u 1+u2

=2

du

=

1 u11

1 u+

1+udu= ln|1 u|+ln|1+u|+C

=ln 1+u 1 u

+C

1+u2=ln +2u 1 u +C

=ln 1+u2

2u 1 u2+1 u2

+C

=ln|secx+tanx|+C.

ؽȩúªµ(1)

11x

(x2+1)2dx=2arctanx+2(x2+1)

+C.

(2)

x2

(x2+1)2dx=12arctanx x2(x2+1)+C.(3) x(x+1)dx=1

12(x+1)d(x2

+1)= 12(x+1)+C.

(4)

x3

(x2+dx

= 1 1)2

12x2dx2

+1

=

x22(x+1)+

x

x+1dx= x2+1ln(x22(x+1)2+1)+C.(1) y²:

6

(0.6)

(0.7)

(0.8)(0.9)

(0.10)