Câu 4: Tính tích phân
$\int_{4}^{6}\sqrt{\frac{x-4}{(x+2)^3}}dx$
$ I$= $\int_{4}^{6}\sqrt{\frac{x-4}{(x+2)^3}}dx$
=$\frac{1}{x+2}\sqrt{\frac{x-4}{x+2}}dx$
Đặt $t=\sqrt{\frac{x-4}{x+2}}\Rightarrow x=\frac{2t^2+4}{1-t^2}\Rightarrow dx=\frac{12t}{(1-t^2)^2}dt$
Ta có: $\frac{1}{x+2}=\frac{1-t^2}{6}$
$I=\int_{0}^{\frac{1}{2}}\frac{1-t^2}{6}.\frac{12t}{(1-t^2)^2}tdt$
$=\int_{0}^{\frac{1}{2}}\frac{2t^2}{1-t^2}dt=\int_{0}^{\frac{1}{2}}(-2-\frac{2}{t^2-1})dt=\int_{0}^{\frac{1}{2}}(-2-\frac{1}{t-1}+\frac{1}{t+1})dt$
=$=(-2x-\ln\left | \frac{t-1}{t+1} \right |)\oint_{0}^{\frac{1}{2}}=\ln3-1$
- hoangmanhquan yêu thích