题目
注 类似地，求int(dx)/(1+x^6). → ((1)/(2sqrt(3))ln|(x^2+sqrt(3)x+1)/(x^2)-sqrt(3)x+1|+(1)/(2)arctanx+(1)/(6)arctanx^3+C)
注 类似地，求$\int\frac{dx}{1+x^{6}}.$ → $\left(\frac{1}{2\sqrt{3}}ln\left|\frac{x^{2}+\sqrt{3}x+1}{x^{2}-\sqrt{3}x+1}\right|+\frac{1}{2}arctanx+\frac{1}{6}arctanx^{3}+C\right)$
题目解答
答案
将被积函数分解为部分分式：
\[
\frac{1}{1+x^6} = \frac{A}{x^2+1} + \frac{Bx+C}{x^2+x\sqrt{3}+1} + \frac{Dx+E}{x^2-x\sqrt{3}+1}
\]
求解系数得：
\[
A = \frac{1}{3}, \quad B = D = 0, \quad C = E = -\frac{1}{3}
\]
积分得：
\[
\int \frac{dx}{1+x^6} = \frac{1}{3} \arctan x - \frac{1}{4\sqrt{3}} \ln \left| \frac{x^2 - x\sqrt{3} + 1}{x^2 + x\sqrt{3} + 1} \right| + C
\]
或表示为：
\[
\boxed{\frac{1}{2\sqrt{3}} \ln \left| \frac{x^2 + x\sqrt{3} + 1}{x^2 - x\sqrt{3} + 1} \right| + \frac{1}{2} \arctan x + \frac{1}{6} \arctan x^3 + C}
\]