问题及解答

求

\[\int\frac{x^2+1}{x^4+1}dx\]

用类似的方法, 求

\[\int\frac{x^2-1}{x^4+1}dx\]

\[
\begin{split}
\int\frac{x^2+1}{x^4+1}dx&=\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\\
&=\int\frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^2+2}\\
&=\frac{1}{\sqrt{2}}\arctan\frac{x-\frac{1}{x}}{\sqrt{2}}+C.
\end{split}
\]

\[
\begin{split}
\int\frac{x^2-1}{x^4+1}dx&=\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\\
&=\int\frac{d(x+\frac{1}{x})}{(x+\frac{1}{x})^2-2}\\
&=\frac{1}{2\sqrt{2}}\ln\biggl|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\biggr|+C\\
&=\frac{1}{2\sqrt{2}}\ln\biggl(\frac{x^2-x\sqrt{2}+1}{x^2+x\sqrt{2}+1}\biggr)+C.
\end{split}
\]

注意要善于利用常用不定积分公式