Question

$\int \frac{(x\ln 2+1)}{x(1+x\cdot 2^x{)}^2}dx$ is equal to
(where $C$ is a constant of integration)

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

Answer 1

The correct option is B $\ln \left(\frac{x\cdot 2^x}{1+x\cdot 2^x}\right)+\frac{1}{1+x\cdot 2^x}+C$

Put $x\cdot 2^x=t$
$\Rightarrow dt=2^x(1+x\ln 2)dx$
$\therefore I=\int \frac{dt}{t{(1+t)}^2}=\int \frac{1+t-t}{t{(1+t)}^2}dt=\int (\frac{1}{t}-\frac{1}{1+t}-\frac{1}{{(1+t)}^2})dt$
$=\ln \left|t\right|-\ln |1+t|+\frac{1}{1+t}+C$
$=\ln \left(\frac{t}{1+t}\right)+\frac{1}{1+t}+C=\ln \left(\frac{x\cdot 2^x}{1+x\cdot 2^x}\right)+\frac{1}{1+x\cdot 2^x}+C$