第2个回答 2014-01-11
cos^2x=(1/2)(cos2x+1)=(1/2)[(1-tan^2x)/(1+tan^2x)+1]=1/(1+tan^2x)
sec^2x=1+tan^2x
f'(tanx+1)
=1/(tan^2x+1)+1+tan^2x
所以:
f'(x)=1/(x^2+1)+(x^2+1)
∫f'(x)dx=∫dx/(1+x^2)+∫(1+X^2)dx
则:
f(x)=arctanx+x+x^3/3+c
f(1)=arctan1+1+1/3+c=2,所以c=2/3-π/4
即f(x)=arctanx+x+x^3+2/3-π/4.