高等数学A上册习题解答 至微分中值定理(14)

华中师范大学专用

3secttantdt,

√

=|3tant|.

√

dx=

3|tant|3sect3secttantdt

=3

|tant|tantdt =3

tan2tdt,0<t<π,

3

tan2tdt,π<t<π.=

3(tant t)+C,0<t<π, 3(tant t)+C,π<t<π.√=√ 3arctan√

+C, 3arctan√x>3,

+C,x< 3.

= √

x2 9 3arctan

3

+C. {2.-u=√

K,du=xdx √

=

xdxx2 9=

x2xdxu2

du

=

u2+9[1 9

u2+9

]du

=u 3arctanu

3+C

= √

x2 9 3arctan3+C.

(9)

dx

1+dx=u==√

====

udu1+u

=u ln(1+u)=√

+ ln(1+√C

+C.

14

(0.34)

(0.35)

(0.36)