Question

$\int \frac{dx}{\left[x(x+1)\right]}=$

• A. $\log \left|\frac{(x+1)}{\left(x\right)}\right|+c$
• B. $\text{log}\left|\frac{\left(x\right)}{(x+1)}\right|+c$
• C. $\log \left|\frac{(x-1)}{\left(x\right)}\right|+c$
• D. $\log \left|\frac{(x–1)}{(x+1)}\right|+c$

Answer 1

The correct option is B

$\text{log}\left|\frac{\left(x\right)}{(x+1)}\right|+c$

Explanation for the correct option:

Finding the integral:

Add and subtract $x$ in the numerator.

$\int \frac{dx}{\left[x(x+1)\right]}$

$\begin{array}{rcl}& =& \int \left(\frac{(x+1-x)dx}{x(x+1)}\right)\\ & =& \int \left(\frac{x+1}{x(x+1)}-\frac{x}{x(x+1)}\right)dx\\ & =& \int \left(\frac{1}{x}-\frac{1}{(x+1)}\right)\mathrm{dx}\\ & =& \log \left|x\right|-\log \left|x+1\right|+c\left[\because \int \left(\frac{1}{x}\right)=\log \left|x\right|\mathrm{similarly}\int \left(\frac{1}{x+1}\right)=\log \left|x+1\right|\right]\\ & =& \log \left|\frac{x}{x+1}\right|+c\left[\because \log \left(\frac{M}{N}\right)=\log \left(M\right)-\log \left(N\right)\right]\end{array}$

Hence the correct answer is option (B).