Tính tích phân $\int\limits^1_{-1} {\frac{x}{x^2+1+x}} \, dx$

Đáp án:

$I = \dfrac12\ln3 - \dfrac{\pi\sqrt3}{6}$

Giải thích các bước giải:

$\begin{array}{l} \quad I = \displaystyle\int\limits_{-1}^1\dfrac{x}{x^2 + x + 1}dx\\ \Leftrightarrow I =  \displaystyle\int\limits_{-1}^1\left[\dfrac{2x +1}{2(x^2 + x +1)} - \dfrac{1}{2(x^2  +x + 1)}\right]dx\\ \Leftrightarrow I = \dfrac12 \displaystyle\int\limits_{-1}^1\dfrac{2x+1}{x^2 +x + 1}dx - \dfrac12\displaystyle\int\limits_{-1}^1\dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}dx\\ \Leftrightarrow I = \dfrac12\ln(x^2 + x + 1)\Bigg|_{-1}^1 - \dfrac12\cdot \dfrac{2}{\sqrt3}\cdot \arctan\left(\dfrac{x + \dfrac12}{\dfrac{\sqrt3}{2}}\right)\Bigg|_{-1}^1\\ \Leftrightarrow I = \dfrac12\ln(x^2 + x + 1)\Bigg|_{-1}^1 - \dfrac{1}{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}\right)\Bigg|_{-1}^1\\ \Leftrightarrow I = \dfrac12\ln3 - \dfrac{\pi\sqrt3}{6} \end{array}$