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: Understanding the Concept
Put \(t=e^{x}\) so that \(dx=\dfrac{dt}{t}\). The integral becomes a rational function of \(t\) that can be split into partial fractions.

Step 2: Substitution
\(\int\dfrac{dt}{t(2t^{2}+3t+1)}=\int\dfrac{dt}{t(2t+1)(t+1)}\).

Step 3: Partial fractions
Write \(\dfrac{1}{t(2t+1)(t+1)}=\dfrac{A}{t}+\dfrac{B}{2t+1}+\dfrac{C}{t+1}\).
Putting \(t=0,-\dfrac{1}{2},-1\) gives \(A=1,\ B=-4,\ C=1\).

Step 4: Integrate and return to x
\(\int\left(\dfrac{1}{t}-\dfrac{4}{2t+1}+\dfrac{1}{t+1}\right)dt=\log t-2\log(2t+1)+\log(t+1)+c\).
With \(t=e^{x}\) this is \(x+\log(e^{x}+1)-2\log(2e^{x}+1)+c\).

Final Answer:
The integral equals x + log(e^x + 1) - 2 log(2e^x + 1) + c. \[ \boxed{\text{(A) }x+\log(e^{x}+1)-2\log(2e^{x}+1)+c} \]