Toán Lớp 12: Tính tích phân $\int\limits^1_{-1} {\frac{x}{x^2+1+x}} \, dx$
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Toán Lớp 12: Tính tích phân $\int\limits^1_{-1} {\frac{x}{x^2+1+x}} \, dx$
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\quad I = \displaystyle\int\limits_{-1}^1\dfrac{x}{x^2 + x + 1}dx\\
\Leftrightarrow I = \displaystyle\int\limits_{-1}^1\left[\dfrac{2x +1}{2(x^2 + x +1)} – \dfrac{1}{2(x^2 +x + 1)}\right]dx\\
\Leftrightarrow I = \dfrac12 \displaystyle\int\limits_{-1}^1\dfrac{2x+1}{x^2 +x + 1}dx – \dfrac12\displaystyle\int\limits_{-1}^1\dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}dx\\
\Leftrightarrow I = \dfrac12\ln(x^2 + x + 1)\Bigg|_{-1}^1 – \dfrac12\cdot \dfrac{2}{\sqrt3}\cdot \arctan\left(\dfrac{x + \dfrac12}{\dfrac{\sqrt3}{2}}\right)\Bigg|_{-1}^1\\
\Leftrightarrow I = \dfrac12\ln(x^2 + x + 1)\Bigg|_{-1}^1 – \dfrac{1}{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}\right)\Bigg|_{-1}^1\\
\Leftrightarrow I = \dfrac12\ln3 – \dfrac{\pi\sqrt3}{6}
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