2 câu trả lời
$I=\displaystyle\int x\ln(3-x^2) \, dx\\ u=\ln(3-x^2) \Rightarrow du=\dfrac{-2x}{3-x^2}\\ dv=xdx \Rightarrow v=\dfrac{x^2-3}{2}\\ I=\ln(|3-x^2|)\dfrac{x^2-3}{2}-\displaystyle\int \dfrac{x^2-3}{2}.\dfrac{-2x}{3-x^2} \, dx\\ =\ln(|3-x^2|)\dfrac{x^2-3}{2}-\displaystyle\int x \, dx\\ =\ln(|3-x^2|)\dfrac{x^2-3}{2}-\dfrac{x^2}{2}+C$
Câu hỏi trong lớp
Xem thêm
