|
|
Ta có: $\int{x\sqrt{(x^2+4)^3}dx}=\frac{1}{2}\int{(x^2+4)^{\frac{3}{2}}dx^2}=\frac{1}{2}.\frac{2}{5}(x^2+4)^{\frac{5}{2}}+C=\frac{1}{5}(x^2+4)^{\frac{5}{2}}+C$.
Do đó: $I=\frac{1}{5}\left( 8^{\frac{5}{2}}-4^{\frac{5}{2}} \right) =\frac{128\sqrt{2}-32}{5}$.
|