積分問題bot107

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問題

\(\displaystyle\int_{1}^{2}\displaystyle\frac{x^2+x+1}{x^2-x+1}dx\)

計算

\(\displaystyle\int_{1}^{2}\displaystyle\frac{x^2+x+1}{x^2-x+1}dx\)

\(=\displaystyle\int_{1}^{2}\displaystyle\frac{x^2-x+1+2x}{x^2-x+1}dx\)

\(=\displaystyle\int_{1}^{2} \biggl(1+\displaystyle\frac{2x}{x^2-x+1}\biggr)dx\)

\(=\displaystyle\int_{1}^{2} \biggl[1+\displaystyle\frac{2x-1}{x^2-x+1}+\displaystyle\frac{1}{\biggl(x-\displaystyle\frac{1}{2}\biggr)^2+\displaystyle\frac{3}{4}}\biggr]dx\)

\(=\biggl[x+\log(x^2-x+1)\biggr]_{1}^{2}+\displaystyle\int_{1}^{2}\displaystyle\frac{1}{\biggl(x-\displaystyle\frac{1}{2}\biggr)^2+\displaystyle\frac{3}{4}}dx\)

\(=1+\log 3+\displaystyle\int_{1}^{2}\displaystyle\frac{1}{\biggl(x-\displaystyle\frac{1}{2}\biggr)^2+\displaystyle\frac{3}{4}}dx\)

第三項

\(x-\displaystyle\frac{1}{2}=\displaystyle\frac{\sqrt{3}}{2}\tan\theta\) とおく。

積分範囲は\(\displaystyle\frac{\pi}{6}\leq \theta\leq\displaystyle\frac{\pi}{3}\)になる。

\(\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{1}{\displaystyle\frac{3}{4}\tan^2\theta+\displaystyle\frac{3}{4}}\cdot\displaystyle\frac{\sqrt{3}}{2\cos^2\theta}d\theta\)

\(=\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{2}{\sqrt{3}} d\theta\)

\(=\displaystyle\frac{\pi}{3\sqrt{3}}\)

答え

\(1+\log 3+\displaystyle\frac{\pi}{3\sqrt{3}}\)

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