高等数学A上册习题解答 至微分中值定理(7)

华中师范大学专用

{1:(n {)x=tant

1

(x2+1)2dx

sec2=tsec4tdt

=

cos2tdt= 12[1+cos2t]dt

=1

2[t+12sin2t]=1x2arctanx+2(x2+1)+C. {2:(©ÜÈ©{)

1

(x2+1)dx

=

2 12xd1x+1

=

112xx+1

1 1

x+1d2x

= 1

12x(x2+1) 2(x2+1)x2

dx=

11

2x(x2+1) 12x2 1

x2+1

dx= 11 12x(x2+1) 2[x arctanx]+C=12arctanx+x2(x+1)+C.(2) y²:

7

(0.11)

(0.12)