Cho \(\int\limits_{\frac{1}{3}}^{1}{\frac{x}{3x+\s...

A

- Hướng dẫn giải

Phương pháp giải:

Trục căn thức ở mẫu.

Giải chi tiết:

\(\begin{align}  I=\int\limits_{\frac{1}{3}}^{1}{\frac{x}{3x+\sqrt{9{{x}^{2}}-1}}dx=}\int\limits_{\frac{1}{3}}^{1}{\frac{x\left( 3x-\sqrt{9{{x}^{2}}-1} \right)}{\left( 3x+\sqrt{9{{x}^{2}}-1} \right)\left( 3x-\sqrt{9{{x}^{2}}-1} \right)}dx=}\int\limits_{\frac{1}{3}}^{1}{\frac{x\left( 3x-\sqrt{9{{x}^{2}}-1} \right)}{1}dx=}\int\limits_{\frac{1}{3}}^{1}{\left( 3{{x}^{2}}-x\sqrt{9{{x}^{2}}-1} \right)dx} \\  =3\int\limits_{\frac{1}{3}}^{1}{{{x}^{2}}dx}-\int\limits_{\frac{1}{3}}^{1}{x\sqrt{9{{x}^{2}}-1}dx} \\ \end{align}\)

Tính \({{I}_{1}}=3\int\limits_{\frac{1}{3}}^{1}{{{x}^{2}}dx}=\left. {{x}^{3}} \right|_{\frac{1}{3}}^{1}=1-\frac{1}{27}=\frac{26}{27}\)

Tính \({{I}_{2}}=\int\limits_{\frac{1}{3}}^{1}{x\sqrt{9{{x}^{2}}-1}dx}=\frac{1}{18}\int\limits_{\frac{1}{3}}^{1}{\sqrt{9{{x}^{2}}-1}d\left( 9{{x}^{2}}-1 \right)}=\frac{1}{27}.\left. \left( 9{{x}^{2}}-1 \right)\sqrt{9{{x}^{2}}-1} \right|_{\frac{1}{3}}^{1}=\frac{1}{27}\left( 16\sqrt{2}-0 \right)=\frac{16\sqrt{2}}{27}\)

\(I={{I}_{1}}-{{I}_{2}}=\frac{26}{27}-\frac{16\sqrt{2}}{27}=a+b\sqrt{2},\,\,\,(a,b\in Q)\Rightarrow a=\frac{26}{27},\,\,b=-\frac{16}{27}\)

Chọn: A