Question

What is $\int \frac{dx}{2^x-1}$ equal to?

• A. $\ln (2^x-1)+c$
• B. $\frac{\ln (1-2^{-x})}{\ln 2}+c$
• C. $\frac{\ln (2^{-x}-1)}{2\ln 2}+c$
• D. $\frac{\ln (1+2^{-x})}{\ln 2}+c$

Answer 1

Answer: (B)

$\frac{\ln (1-2^{-x})}{\ln 2}+c$

$\int \frac{dx}{2^x-1}=\int \frac{2^{-x}dx}{1-2^{-x}}$

Now, put $1-2^{-x}=t⟹dt=2^{-x}\ln 2dx$

$\int \frac{dx}{2^x-1}=\int \frac{1}{\ln 2}\frac{2^{-x}\ln 2dx}{1-2^{-x}}=\frac{1}{\ln 2}\int \frac{dt}{t}=\frac{1}{\ln 2}\ln t+c=\frac{\ln (1-2^{-x})}{\ln 2}+c$