Question

If $\int (e^{2x}+2e^x-e^{-x}-1)e^{\left(e^x+e^{-x}\right)}dx=g\left(x\right)e^{\left(e^x+e^{-x}\right)}+C$, where $C$ is a constant, then $g\left(0\right)$ is equal to

• A. $2$
• B. $e$
• C. $1$
• D. $e^2$

Answer 1

The correct option is A

$2$

Explanation for the correct option.

Find the value of $g\left(0\right)$:

Given,

$\int (e^{2x}+2e^x-e^{-x}-1)e^{\left(e^x+e^{-x}\right)}dx=g\left(x\right)e^{\left(e^x+e^{-x}\right)}+C$.

Now, Differentiating both side,

$\begin{array}{rcl}(e^{2x}+2e^x-e^{-x}-1)e^{\left(e^x+e^{-x}\right)}& =& \frac{d}{dx}\left(g\left(x\right)e^{\left(e^x+e^{-x}\right)}\right)\\ \Rightarrow (e^{2x}+2e^x-e^{-x}-1)e^{\left(e^x+e^{-x}\right)}& =& g\text{'}\left(x\right)\left[e^{\left(e^x+e^{-x}\right)}\right]+g\left(x\right)\left[e^x-e^{-x}\right]\left[e^{\left(e^x+e^{-x}\right)}\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}\mathbf{=}\mathbf{u}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{v}\mathbf{+}\mathbf{v}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{u}\mathbf{]}\\ & & \\ \Rightarrow (e^{2x}+2e^x-e^{-x}-1)e^{\left(e^x+e^{-x}\right)}& =& \left[e^{\left(e^x+e^{-x}\right)}\right]\left[g\text{'}\left(x\right)+g\left(x\right)\left[e^x-e^{-x}\right]\right]\\ \Rightarrow (e^{2x}+2e^x-e^{-x}-1)& =& \left[g\text{'}\left(x\right)+g\left(x\right)\left[e^x-e^{-x}\right]\right]\\ \Rightarrow \left(e^{2x}-1\right)+e^x+e^x-e^{-x}& =& g\text{'}\left(x\right)+g\left(x\right)\left[e^x-e^{-x}\right]\\ \Rightarrow e^x\left(e^x-e^{-x}\right)+\left(e^x-e^{-x}\right)+e^x& =& g\text{'}\left(x\right)+g\left(x\right)\left[e^x-e^{-x}\right]\\ \Rightarrow \left(e^x-e^{-x}\right)\left(e^x+1\right)+e^x& =& \left[e^x-e^{-x}\right]g\left(x\right)+g\text{'}\left(x\right)\end{array}$

On comparing the coefficient of both side,

$g\left(x\right)=\left(e^x+1\right)$

Therefore, $g\left(0\right)=2$.

Hence, the correct option is A.