Tính tích phân $\displaystyle \int_{0}^{1}\frac{\ln(1+x^2)}{1+x}dx$ What i have try: Let $\displaystyle I(a) = \int_{0}^{1}\frac{\ln(1+ax^2)}{1+x^2}\;,$ Then $\displaystyle I'(a) = \int_{0}^{1}\frac{x^2}{(1+ax^2)(1+x)}dx$ So $\displaystyle I'(a) = -\frac{1}{1+a}\int_{0}^{1}\left[\frac{1-x}{1+ax^2}-\frac{1}{1+x}\right]dx$ [RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT] thay đổi nội dung bởi: man111, 02-07-2016 lúc 09:30 PM |