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Cho hàm số \[f\left( x \right) = \frac{{{2^{x + 1}} - {5^{x - 1}}}}{{{{10}^x}}}\]. Khi đó:

\[\int {f\left( x \right){\rm{d}}x} = \frac{{{5^x}}}{{2\ln 5}} - \frac{{5 \cdot {2^x}}}{{\ln 2}} + C\].

\[\int {f\left( x \right){\rm{d}}x} = \frac{2}{{{5^x} \cdot \ln 5}} - \frac{1}{{5 \cdot {2^x} \cdot \ln 2}} + C\].

\[\int {f\left( x \right){\rm{d}}x} = - \frac{{{5^x}}}{{2\ln 5}} + \frac{{5 \cdot {2^x}}}{{\ln 2}} + C\].

\[\int {f\left( x \right){\rm{d}}x} = - \frac{2}{{{5^x} \cdot \ln 5}} + \frac{1}{{5 \cdot {2^x} \cdot \ln 2}} + C\].

Giải thích

Ta có: \[f\left( x \right) = \frac{{{2^{x + 1}} - {5^{x - 1}}}}{{{{10}^x}}} = \frac{{2 \cdot {2^x} - \frac{1}{5} \cdot {5^x}}}{{{{10}^x}}} = 2 \cdot {\left( {\frac{1}{5}} \right)^x} - \frac{1}{5} \cdot {\left( {\frac{1}{2}} \right)^x}\]

\[ \Rightarrow \]\[\int {f\left( x \right){\rm{d}}x} = \int {\left[ {2 \cdot {{\left( {\frac{1}{5}} \right)}^x} - \frac{1}{5} \cdot {{\left( {\frac{1}{2}} \right)}^x}} \right]{\rm{d}}x} = 2 \cdot \frac{{{{\left( {\frac{1}{5}} \right)}^x}}}{{\ln \left( {\frac{1}{5}} \right)}} - \frac{1}{5} \cdot \frac{{{{\left( {\frac{1}{2}} \right)}^x}}}{{\ln \left( {\frac{1}{2}} \right)}} = - \frac{2}{{{5^x} \cdot \ln 5}} + \frac{1}{{5 \cdot {2^x} \cdot \ln 2}} + C\]. Chọn D.