Xét \(\int\limits_{0}^{1}{{{x}^{3}}{{\left( {{x}^{2}}+1 \right)}^{2021}}}\text{d}x\), nếu đặt \(u={{x}^{2}}+1\) thì \(\int\limits_{0}^{1}{{{x}^{3}}{{\left( {{x}^{2}}+1 \right)}^{20...

* Hướng dẫn giải

Xét \(I=\int\limits_{0}^{1}{{{x}^{3}}{{\left( {{x}^{2}}+1 \right)}^{2021}}}\text{d}x\).

Đặt \({{x}^{2}}+1=u\Rightarrow {{x}^{2}}=u-1\).

Ta có \(2x\text{d}x=\text{d}u\Rightarrow x\text{d}x=\frac{\text{d}u}{2}\).

Đổi cận:

\(\begin{align} & x=0\Rightarrow u=1 \\ & x=1\Rightarrow u=2 \\ \end{align}\)

Vậy \(I=\frac{1}{2}\int\limits_{1}^{2}{\left( u-1 \right){{u}^{2021}}\text{d}u}\).