这道数学题求解答

设 f(x)=∑{[x^(4n+1)]/(4n+1)}，则f(0)=0f'(x)=∑{[x^(4n+1)]/(4n+1)]}'=∑x^(4x)是等比数列，首项a(1)=x^4，公比q=x^4。∴ f'(x)=(x^4)/(1-x^4)(x^4)/(1-x^4)=-1+1/(1-x^4)1/(1-x^4)=0.5[1/(1-x²)+1/(1+x²)]1/(1-x²)=0.5[1/(1-x)+1/(1+x)]f(x)=∫【x，0】(x^4)dx/(1-x^4)=∫[-1+0.25/(1-x)+0.25/(1+x)+0.5/(1+x²)]dx=-x-0.25ln|1-x|+0.25ln|1+x|+0.5arctanx