MCQ
જો ${I_1} = \int_e^{{e^2}} {\frac{{dx}}{{\log x}}} $ અને ${I_2} = \int_1^2 {\frac{{{e^x}}}{x}\,dx,} $ તો
- ✓${I_1} = {I_2}$
- B${I_1} > {I_2}$
- C${I_1} < {I_2}$
- Dએકપણ નહીં.
so that $dx = x\,du = {e^u}du$
Also as $x = e$ to ${e^2},u = 1$ to $2$
Thus, ${I_1} = \int_1^2 {\frac{{{e^u}}}{u}du = \int_1^2 {\frac{{{e^x}}}{x}dx} } $.
Hence, ${I_1} = {I_2}$.
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