Question

$\int \left(\frac{e^x-1}{e^x+1}\right)dx=f\left(x\right)+c,thenf\left(x\right)=$

• A. $2\log \left(e^x+1\right)$
• B. $2\log \left|e^{2x}–1\right|$
• C. $2\log \left|e^x+1\right|–x$
• D. $2\log \left(e^{2x}+1\right)$

Answer 1

The correct option is C

$2\log \left|e^x+1\right|–x$

Explanation for the correct option:

Finding the given integral:

Given $\int \left(\frac{e^x-1}{e^x+1}\right)dx=f\left(x\right)+c$

Let us consider

$$\begin{gathered} u=e^x \\ du=e^{x}dx \\ \Rightarrow \frac{\mathrm{du}}{{\mathrm{e}}^{\mathrm{x}}}=\mathrm{dx} \\ \Rightarrow \frac{\mathrm{du}}{\mathrm{u}}=\mathrm{dx} \end{gathered}$$

Substituting in the given integral

$\begin{array}{rcl}& =& \int \frac{u-1}{u(u+1)}du\\ & =& \int \frac{2u-(u+1)}{u(u+1)}du\left[\text{substituting}u=2u-u\right]\\ & =& 2\int \frac{1}{u+1}du-\int \frac{1}{u}du\\ & =& 2\ln \left|u+1\right|-\ln \left|u\right|+C\left[\int \frac{1}{u+1}du=\ln \left|u+1\right|\right]\\ & =& 2\ln \left|e^x+1\right|-x+C\left[\text{Resubstituting}u=e^x\right]\end{array}$

Hence the correct answer is option (C).