Question

Prove that ${\int }_{log2}^x\frac{1}{e^x-1}dx=log\frac{3}{2}$ when $x=\log 4$.

Answer 1

Let $I={\int }_{\log 2}^x\frac{1}{e^x-1}dt$

Put $e^x=t$

$\Rightarrow e^{x}dx=dt\Rightarrow dx=\frac{dt}{e^x}=\frac{dt}{t}$

When $x=\log 2\Rightarrow t=2$

When $x=\log 4\Rightarrow t=e^{\log 4}=4$

So, $I={\int }_2^4\frac{1}{t(t-1)}dt$

Resolving $\frac{1}{t(t-1)}$ into partial fractions

$\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$

$\Rightarrow 1=A(t-1)+Bt$

Put $t=1$

$\Rightarrow B=1$

Put $t=0$

$\Rightarrow A=-1$

$\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$

Integrating both sides, we get

$={\int }_2^4(\frac{1}{t-1}-\frac{1}{t})dt$
$={[\log (t-1)-\log t]}_2^4$

$=\log 3-\log 1-\log 4+\log 2=\log 3-2\log 2+\log 2=\log 3-\log 2$

$=\log \frac{3}{2}$

Hence Proved