Lời giải của Tự Học 365

Ta có \(I=\int\limits_{0}^{1}{\frac{x}{\sqrt{3x+1}+\sqrt{2x+1}}\,\text{d}x}=\int\limits_{0}^{1}{\frac{x\left( \sqrt{3x+1}-\sqrt{2x+1} \right)}{{{\left( \sqrt{3x+1} \right)}^{2}}-{{\left( \sqrt{2x+1} \right)}^{2}}}\,\text{d}x}\)

\(\begin{array}{l}
= \int\limits_0^1 {\frac{{x\left( {\sqrt {3x + 1} - \sqrt {2x + 1} } \right)}}{{3x + 1 - 2x - 1}}} = \int\limits_0^1 {\left( {\sqrt {3x + 1} - \sqrt {2x + 1} } \right){\rm{d}}x} .\\
= \left. {\left( {\frac{1}{3}.\frac{{\sqrt {{{\left( {3x + 1} \right)}^3}} }}{{\frac{3}{2}}} - \frac{1}{2}.\frac{{\sqrt {{{\left( {2x + 1} \right)}^3}} }}{{\frac{3}{2}}}} \right)} \right|_0^1 = \left. {\left( {\frac{2}{9}.\sqrt {{{\left( {3x + 1} \right)}^3}} - \frac{1}{3}.\sqrt {{{\left( {2x + 1} \right)}^3}} } \right)} \right|_0^1\\
= \left. {\frac{1}{9}\left[ {2\sqrt {{{\left( {3x + 1} \right)}^3}} - 3\sqrt {{{\left( {2x + 1} \right)}^3}} } \right]} \right|_0^1 = \frac{1}{9}\left( {16 - 9\sqrt 3 + 1} \right) = \frac{{17 - 9\sqrt 3 }}{9} \Rightarrow \left\{ \begin{array}{l}
a = 17\\
b = - \,9
\end{array} \right..
\end{array}\)

Vậy \(T=a+b=17-9=8.\)

Đáp án cần chọn là: d