Đặt $u = \ln(x^2+x+1) \ ; dv = \frac{dx}{(x+1)^2}$
$du = \frac{2x+1}{x^2+x+1}dx \ ; v = -\frac{1}{x+1}$
$I = -\left.\ln(x^2+x+1)\frac{1}{x+1}\right|_{0}^{1} + \int\limits_{0}^{1}\frac{2x+1}{(x^2+x+1)(x+1)}dx$
$I = -\frac{1}{2}\ln3+J$
Với J
$J = \int\limits_{0}^{1}\frac{2x+1}{(x^2+x+1)(x+1)}dx$
$J = \int\limits_{0}^{1}\frac{x+2}{(x^2+x+1)}dx-\int\limits_{0}^{1}\frac{1}{(x+1)}dx$
$J = \int\limits_{0}^{1}\frac{x+1/2}{(x^2+x+1)}dx+\int\limits_{0}^{1}\frac{3/2}{(x^2+x+1)}dx-\int\limits_{0}^{1}\frac{1}{(x+1)}dx$
$J = \frac{1}{2}\int\limits_{0}^{1}\frac{d(2x+1)}{(x^2+x+1)}dx+\frac{3}{2}\int\limits_{0}^{1}\frac{1}{((x+1/2)^2+3/4)}dx-\int\limits_{0}^{1}\frac{1}{(x+1)}dx$
$J = \left.\ln(x^2+x+1)\right|_0^1 +J_1 -\left.\ln|x+1|\right|_0^1=$
$J = \ln(3) +\frac{3}{2}J_1 -\ln 2$
Với $J_2 = \int\frac{1}{((x+1/2)^2+3/4)}dx$ (chút nữa thay cận vào là xong)
Đặt $x+1/2 = \frac{\sqrt{3}}{2}\tan t \to dx = \frac{\sqrt{3}}{2}\frac{1}{\cos^2 t}dt =\frac{\sqrt{3}}{2}(1+\tan^2t)dt$
$J_2 = \int\frac{\sqrt{3}/2(1+\tan^2t)}{3/4(1+\tan^2t)}dt=\frac{2}{\sqrt 3}t+C = \frac{2}{\sqrt 3}\arctan (x+1/2)+C$
$J_1 = \left.\frac{2}{\sqrt 3}\arctan (x+1/2)\right|_0^1=\frac{2}{\sqrt 3}(\arctan (3/2) - \arctan(1/2))$
Vậy $I = -\frac{1}{2}\ln 3+ \ln 3 - \ln 2 +\frac{3}{2}\frac{2}{\sqrt 3}(\arctan (3/2) - \arctan(1/2))$
hay
$I = \frac{1}{2}\ln 3 - \ln 2+\sqrt{3}(\arctan (3/2) - \arctan(1/2))$
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