Tính tích phân

N

nguyenbahiep1

$I = \int_{0}^{1}( x - \frac{x}{(x+1)(x^2-x+1)})dx \\ \\ I = 1 - \frac{1}{3}\int_{0}^{1}(\frac{x+1}{x^2-x+1}-\frac{1}{x+1})dx \\ \\ I = 1 + \frac{1}{3}ln|x+1| \big|_0^1 - \frac{1}{3}\int_{0}^{1}\frac{x+1}{x^2-x+1}dx$

Ta chỉ cần tính

$I_1 = \int_{0}^{1}\frac{x+1}{x^2-x+1}dx = \frac{1}{2} \int_{0}^{1}\frac{2x-1}{x^2-x+1}dx + \frac{1}{2}\int_{0}^{1}\frac{3}{x^2-x+1}dx \\ \\ I_1 = \frac{1}{2}ln|x^2-x+1| \big|_0^1 \\ \\ I_2 = \int_{0}^{1}\frac{1}{x^2-x+1}dx = \int_{0}^{1}\frac{1}{(x-\frac{1}{2})^2+ (\frac{\sqrt{3}}{2})^2}dx \\ \\ x- \frac{1}{2} = \frac{\sqrt{3}}{2}tant$
 
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