複雑な2重根号を含む定積分

\[ \sqrt{x^{2}+1+\sqrt{x^{4}+x^{2}+1}}=\sqrt{a}+\sqrt{b} \] とおくと、
\[ x^{2}+1+\sqrt{x^{4}+x^{2}+1}=a+b+2\sqrt{ab} \] \[ \begin{cases} x^{2}+1=a+b\\ x^{4}+x^{2}+1=4ab \end{cases} \] これより、
\begin{align*} 4ab & =x^{4}+x^{2}+1\\ & =\left(x^{2}+1\right)^{2}-x^{2}\\ & =\left(a+b\right)^{2}-x^{2} \end{align*} となり、\(x^{2}\)について解くと、
\begin{align*} x^{2} & =\left(a+b\right)^{2}-4ab\\ & =\left(a-b\right)^{2} \end{align*} となるので、
\[ \pm x=a-b \] となる。
従って、
\[ \begin{cases} a+b=x^{2}+1\\ a-b=\pm x \end{cases} \] となるので、
\[ \begin{cases} a=\frac{1}{2}\left(x^{2}\pm x+1\right)\\ b=\frac{1}{2}\left(x^{2}\mp x+1\right) \end{cases} \] となり\(a,b\)の対称性より、
\[ \begin{cases} a=\frac{1}{2}\left(x^{2}+x+1\right)\\ b=\frac{1}{2}\left(x^{2}-x+1\right) \end{cases} \] とする。
これより元の積分は、
\[ \int\sqrt{x^{2}+a}dx=\frac{x}{2}\sqrt{x^{2}+a}+\frac{a}{2}\log\left(\sqrt{x^{2}+a}+x\right)+C \] を使うと、
\begin{align*} \int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^{2}+1+\sqrt{x^{4}+x^{2}+1}}dx & =\int_{-\frac{1}{2}}^{\frac{1}{2}}\left(\sqrt{a}+\sqrt{b}\right)dx\\ & =\int_{-\frac{1}{2}}^{\frac{1}{2}}\left(\sqrt{\frac{1}{2}\left(x^{2}+x+1\right)}+\sqrt{\frac{1}{2}\left(x^{2}-x+1\right)}\right)dx\\ & =\frac{\sqrt{2}}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\left(\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right)dx\\ & =\frac{\sqrt{2}}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\left(\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right)dx\\ & =\frac{\sqrt{2}}{2}\left[\frac{1}{2}\left(x+\frac{1}{2}\right)\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}+x+\frac{1}{2}\right)+\frac{1}{2}\left(x-\frac{1}{2}\right)\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}+x-\frac{1}{2}\right)\right]_{-\frac{1}{2}}^{\frac{1}{2}}\\ & =\frac{\sqrt{2}}{2}\left[\frac{1}{2}\sqrt{1+\frac{3}{4}}+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{1+\frac{3}{4}}+1\right)+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{\frac{3}{4}}\right)-\left\{ \frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{\frac{3}{4}}\right)-\frac{1}{2}\sqrt{1+\frac{3}{4}}+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{1+\frac{3}{4}}-1\right)\right\} \right]\\ & =\frac{\sqrt{2}}{2}\left\{ \sqrt{1+\frac{3}{4}}+\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{1+\frac{3}{4}}+1\right)-\frac{1}{2}\cdot\frac{3}{4}\log\left(\sqrt{1+\frac{3}{4}}-1\right)\right\} \\ & =\frac{\sqrt{2}}{2}\left\{ \frac{\sqrt{7}}{2}+\frac{3}{8}\log\left(\frac{\sqrt{7}}{2}+1\right)-\frac{3}{8}\log\left(\frac{\sqrt{7}}{2}-1\right)\right\} \\ & =\frac{\sqrt{2}}{2}\left\{ \frac{\sqrt{7}}{2}+\frac{3}{8}\log\left(\frac{\sqrt{7}+2}{\sqrt{7}-2}\right)\right\} \\ & =\frac{\sqrt{2}}{2}\left\{ \frac{\sqrt{7}}{2}+\frac{3}{8}\log\left(\frac{11+4\sqrt{7}}{3}\right)\right\} \\ & =\frac{\sqrt{14}}{4}+\frac{3\sqrt{2}}{16}\log\left(\frac{11+4\sqrt{7}}{3}\right) \end{align*}