Question

What is $\int {\dfrac{{dx}}{{{2^x} - 1}}} $ equal to?
A. $\ln ({2^x} - 1) + c$
B. \[\dfrac{{\ln (1 - {2^{ - x}})}}{{\ln 2}} + c\]
C. \[\dfrac{{\ln ({2^{ - x}} - 1)}}{{\ln 2}} + c\]
D. \[\dfrac{{\ln (1 + {2^{ - x}})}}{{\ln 2}} + c\]

Answer 1

Hint: Here, we need to solve the given integral by substitution method and then simplifying the given integral and then the use of the required formulae of integration.

Complete step-by-step answer:
Let $I = \int {\dfrac{{dx}}{{{2^x} - 1}}} $
Let ${2^x} - 1 = t$
Differentiate it w.r.t x
$
   \Rightarrow \dfrac{{d({2^x} - 1)}}{{dx}} = \dfrac{{dt}}{{dx}} \\
   \Rightarrow \ln 2 \times {2^x}dx = dt \\
   \Rightarrow dx = \dfrac{{dt}}{{{2^x}\ln 2}}{\text{ & }}{{\text{2}}^x} = t + 1 \\
   \Rightarrow dx = \dfrac{{dt}}{{(t + 1)\ln 2}} \\
$
Now substitute this value in the integral
$I = \int {\dfrac{1}{t}\dfrac{{dt}}{{(t + 1)\ln 2}}} $
Now apply partial fraction for $\dfrac{1}{{t(t + 1)}}$
$ = \dfrac{1}{{\ln 2}}\int {\left( {\dfrac{1}{t} - \dfrac{1}{{t + 1}}} \right)} dt$
Now apply integration
$
   = \dfrac{1}{{\ln 2}}[\ln t - \ln (t + 1)] + c \\
   = \dfrac{1}{{\ln 2}}[\ln \dfrac{t}{{t + 1}}] + c \\
$
Substitute the value of $t = {2^x} - 1$
$
   = \dfrac{1}{{\ln 2}}\left[ {\ln \dfrac{{({2^x} - 1)}}{{{2^x}}}} \right] + c \\
   \Rightarrow \dfrac{{\ln (1 - {2^{ - x}})}}{{\ln 2}} + c \\
 $
Therefore, option B is correct.

Note: Integrals can be simplified to standard form by substitution techniques which can be easily evaluated using the standard integration formulas.