Question

\[\int \frac{\mathrm{d} x}{2 \mathrm{e}^{2 x}+3 \mathrm{e}^x+1}=\]

• A. \(x+\mathrm{log}(\mathrm{e}^{x}+1)-2\mathrm{log}(2\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration
• B. \(x-\mathrm{log}(\mathrm{e}^{x}+1)+4\mathrm{log}(\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration
• C. \(x+\mathrm{log}(\mathrm{e}^{x}+1)-4\mathrm{log}(2\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration
• D. \(x-\mathrm{log}(\mathrm{e}^{x}+1)+2\mathrm{log}(2\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration

Hint:

Substitute t = e^x and split the resulting rational function into partial fractions.

Answer 1

The Correct Option is A

Step 1: Factorise the denominator
$2e^{2x}+3e^{x}+1=(2e^{x}+1)(e^{x}+1)$.

Step 2: Differentiate the candidate answer
$\dfrac{d}{dx}\left[x+\log(e^{x}+1)-2\log(2e^{x}+1)\right]=1+\dfrac{e^{x}}{e^{x}+1}-\dfrac{4e^{x}}{2e^{x}+1}$.

Step 3: Combine over a common denominator
The numerator is $(2e^{2x}+3e^{x}+1)+(2e^{2x}+e^{x})-(4e^{2x}+4e^{x})=1$.
So the derivative is $\dfrac{1}{(e^{x}+1)(2e^{x}+1)}$, which is the integrand.

Final Answer:
Differentiating x + log(e^x + 1) - 2 log(2e^x + 1) returns the integrand, so it is the integral. \[ \boxed{\text{(A) }x+\log(e^{x}+1)-2\log(2e^{x}+1)+c} \]