高等数学A上册习题解答 至微分中值定理(21)

华中师范大学专用

1

dxx+1

1

dx=

x4+2x2+1 2x2

1

dx=

(x2+1)2 2x2

1

=dx(x2++1)(x2 x+1)

√ √11 x+x = dxx2++1x2 +1

√√√ √11 + (2x+(2x = dxx2++1x2 x+1√√

√√√√22

[ln(x+x+1) ln(x x+1)]+[arctan(+1)+arctan(x 1)]+C=84√√√

√√x2++1=ln+[arctan(x+1)+arctan( 1)]+C

84x2 +1

(0.57)4.

1

dx3+sinx

2

=dx

7 cos2x

21u=tanx

=======du u1+u2

7 1

du

=

4u+3122u

=arctan+C412tanx=arctan+C.25.

(0.58)

21