Tích phân \(\int\limits^3_0\frac{x^2}{\left(1+x\right)^{\frac{3}{2}}}\text{d}x\) bằng
\(\frac{2}{3}\). \(1\). \(\frac{4}{3}\). \(\frac{5}{3}\). Hướng dẫn giải:Đặt $1 + x = t$ (để mẫu số là đơn thức)
$ \Rightarrow x = t - 1 ; \text{d}x = \text{d}t$. Đổi cận: \(x|^3_0\Rightarrow t|^4_1\)
\(I=\int\limits^4_1\frac{\left(1-t\right)^2}{t^{\frac{3}{2}}}\text{d}t\)
\(=\int\limits^4_1\frac{1-2t+t^2}{t^{\frac{3}{2}}}\text{d}t\)
\(=\int\limits^4_1\left(t^{-\frac{3}{2}}-2t^{-\frac{1}{2}}+t^{\frac{1}{2}}\right)\text{d}t\)
\(=\left[\frac{1}{-\frac{3}{2}+1}t^{-\frac{3}{2}+1}-2.\frac{1}{-\frac{1}{2}+1}t^{-\frac{1}{2}+1}+\frac{1}{\frac{1}{2}+1}t^{\frac{1}{2}+1}\right]|^4_1\)
\(=\left[-2t^{-\frac{1}{2}}-4t^{\frac{1}{2}}+\frac{2}{3}t^{\frac{3}{2}}\right]|^4_1\)
\(=\frac{5}{3}\).