Consider the integral $A={\int }_0^1\frac{e^x-1}{x}\mathrm{}dx$ and $B={\int }_0^1\frac{x}{e^x-1}dx$. Then, which of the following is incorrect?

• A

  $B<1$
• B

  $A>1$
• Correct answer
  C

  $B>A$
• D

  $A>\frac{1}{2}$

Solution

As, $e^x-1>x \forall x\in \left(0,\mathrm{\infty }\right)$
$\therefore \frac{e^x-1}{x}>1$
Hence, $A={\int }_0^1\frac{e^x-1}{x}dx>{\int }_0^{1}1dx$
$\Rightarrow A>1$
And, $B={\int }_0^1\frac{x}{e^x-1}dx<{\int }_0^{1}1dx$
$\Rightarrow B<1$
$\therefore B<1<A$

Hence, the correct answer is option: 'C'.