一个无理分式的不定积分

青青子衿

\[ \int \frac{x^2}{a^2+x^2+\sqrt{a^2+x^2}}{\rm\,d}x \]
\[ \frac{x^2}{a^2+x^2+\sqrt{a^2+x^2}}=1-\frac{a^2-1}{a^2-1+x^2}+\frac{a^2-1}{\left(a^2-1+x^2\right)\sqrt{a^2+x^2}}-\frac{1}{\sqrt{a^2+x^2}} \]\begin\begin{align*}
\int \frac{x^2}{a^2+x^2+\sqrt{a^2+x^2}}{\rm\,d}x=
\begin{cases}
\begin{array}{r}  
\begin{split}  
x+\frac{\sqrt{1-a^2}}{2}\ln\left|\frac{1+\dfrac{x}{\sqrt{\left(1-a^2\right)\left(a^2+x^2\right)}}}{1-\dfrac{x}{\sqrt{\left(1-a^2\right)\left(a^2+x^2\right)}}}\right|\\
-\ln\left(x+\sqrt{a^{\overset{\,}{2}}+x^2}\,\right)-\frac{\sqrt{1-a^2}}{2}\ln\left|\frac{1+\dfrac{x}{\sqrt{1-a^2}}}{1-\dfrac{x}{\sqrt{1-a^2}}}\right|\end{split}  
\end{array}&\quad\,a\in(-1,1)\\
\\\hline\\
\qquad\qquad\qquad\qquad x-\ln\left(x+\sqrt{a^{\overset{\,}{2}}+x^2}\,\right)&\quad\,a=1\\
\\\hline\\
\begin{array}{r}  
\begin{split}  
x+\sqrt{a^2-1}\arctan\left(\dfrac{x}{\sqrt{\left(a^2-1\right)\left(a^2+x^2\right)}}\right)\\
-\ln\left(x+\sqrt{a^{\overset{\,}{2}}+x^2}\,\right)-\sqrt{a^2-1}\arctan\left(\dfrac{x}{\sqrt{a^2-1}}\right)
\end{split}  
\end{array}&\quad\,a\in(-\infty,-1)\cup(1,+\infty)
\end{cases}
\end{align*}

orzweb111

This is horrible. No wonder most students are scared away from mathematics.

hbghlyj

青青子衿 发表于 2018-11-3 08:36
\begin\begin{align*}

这里有两个\begin，多打了一个