Question

$\displaystyle \int \frac{ln \left(x + 1\right) - ln ⁡ x}{x \left(x + 1\right)}dx \, $ is equal to (where $C$ is an arbitarary constant)

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

Answer 1

Answer: (A)

Put $ln \left(x + 1\right)-ln ⁡ x=t$
$\Rightarrow \, \frac{1}{x + 1}-\frac{1}{x}=\frac{d t}{d x}\Rightarrow \, \frac{x - \left(x + 1\right)}{x \left(x + 1\right)}=\frac{d t}{d x}$
$\Rightarrow \, \frac{- d x}{x \left(x + 1\right)}=dt\Rightarrow \, \frac{d x}{x \left(x + 1\right)}=- \, dt$
so question becomes
$- \, \displaystyle \int ‍t \, dt=-\frac{t^{2}}{2}+C$