Lời giải của Tự Học 365

Đặt \(t=\sqrt[3]{1+{{x}^{2}}}\Leftrightarrow {{t}^{3}}=1+{{x}^{2}}\Leftrightarrow 2x\,\text{d}x=3{{t}^{2}}\,\text{d}t\Leftrightarrow x\,\text{d}x=\frac{3{{t}^{2}}}{2}\,\text{d}t\) và \(\left\{\begin{align}  x=0\,\,\Rightarrow \,\,t=1 \\  x=\sqrt{7}\,\,\Rightarrow \,\,t=2 \\ \end{align} \right..\)

Khi đó \(\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{3}}\,\text{d}x}{\sqrt[3]{1+{{x}^{2}}}}}=\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{2}}}{\sqrt[3]{1+{{x}^{2}}}}.x\,\text{d}x}=\int\limits_{1}^{2}{\frac{{{t}^{3}}-1}{t}.\frac{3{{t}^{2}}}{2}\,\text{d}t}=\frac{3}{2}\,\int\limits_{1}^{2}{\left( {{t}^{4}}-t \right)\,\text{d}t}=\frac{141}{20}.\)

Vậy \(\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{3}}\,\text{d}x}{\sqrt[3]{1+{{x}^{2}}}}}=\frac{m}{n}\Rightarrow\left\{ \begin{align}  m=141 \\  n=20 \\ \end{align} \right.\,\,\xrightarrow{{}}\,\,m-7n=141-7.20=1.\)

Đáp án cần chọn là: b