Answer

问题及解答

求不定积分 $\int\frac{x^2+1}{x\sqrt{1+x^4}}dx$

Posted by haifeng on 2016-04-28 19:29:21 last update 2016-04-28 19:31:25 | Edit | Answers (1)

求不定积分

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

 


[Hint] 令 $x^2=\tan t$,   $t\in(0,\frac{\pi}{2})$.

1

Posted by haifeng on 2016-04-29 04:30:23

令 $x^2=\tan t$, 这里 $t\in(0,\frac{\pi}{2})$. 则 $x=\sqrt{\tan t}$, $dx=d\sqrt{\tan t}=\frac{\sec^2 t}{2\sqrt{\tan t}}dt$.

\[
\begin{split}
\int\frac{x^2+1}{x\sqrt{1+x^4}}dx &=\int\frac{\tan t+1}{\sqrt{\tan t}\cdot\sqrt{1+\tan^2 t}}\cdot\frac{\sec^2 t}{2\sqrt{\tan t}}dt\\
&=\int\frac{(1+\tan t)\sec^2 t}{2\tan t\cdot|\sec t|}dt=\int\frac{(1+\tan t)\sec t}{2\tan t}dt\\
&=\int\frac{1+\tan t}{2\tan t\cdot\cos t}dt=\int\frac{1+\tan t}{2\sin t}dt\\
&=\frac{1}{2}\biggl[\int\frac{1}{\sin t}dt+\int\frac{1}{\cos t}dt\biggr]
\end{split}
\]

\[
\int\frac{1}{\sin t}dt=\int\frac{\sin t}{\sin^2 t}dt=\int\frac{-d\cos t}{1-\cos^2},
\]

\[
\int\frac{1}{\cos t}dt=\int\frac{\cos t}{\cos^2 t}dt=\int\frac{d\sin t}{1-\sin^2}.
\]

然后换元即可.

略.