Question

If \(I_{1}=\int \frac{e^{x}}{e^{2x}+e^{x}+1}dx\) and \(I_{2}=\int \frac{1}{e^{-2x}+e^{-x}+1}dx\), then \(I_{1}-I_{2}=\)

• A. \(-\frac{1}{2}\log |e^{2x}+e^{x}+1|+\sqrt{3}\tan ^{-1}\left(\frac{2e^{x}+1}{\sqrt{3}}\right)+c\)
• B. \(-\frac{1}{2}\log |e^{2x}+e^{x}+1|+\frac{\sqrt{3}}{2}\log |\frac{2e^{x}+1-\sqrt{3}}{2e^{x}+1+\sqrt{3}}|+c\)
• C. \(-\frac{1}{2}\log |e^{2x}+e^{x}+1|-\sqrt{3}\tan ^{-1}\left(\frac{2e^{x}+1}{\sqrt{3}}\right)+c\)
• D. \(-\frac{1}{2}\log |e^{2x}+e^{x}+1|-\frac{\sqrt{3}}{2}\log |\frac{2e^{x}+1-\sqrt{3}}{2e^{x}+1+\sqrt{3}}|+c\)

Hint:

Rewrite I2 with e^(2x) in the numerator, subtract, and substitute t = e^x.

Answer 1

The Correct Option is A

Step 1: Differentiate the first option
Let $F=-\dfrac{1}{2}\log (t^{2}+t+1)+\sqrt{3}\tan ^{-1}\dfrac{2t+1}{\sqrt{3}}$ with $t=e^{x}$.

Step 2: Derivative in t
The log term gives $-\dfrac{2t+1}{2(t^{2}+t+1)}$.
The arctangent term gives $\sqrt{3}\cdot \dfrac{2/\sqrt{3}}{1+\frac{(2t+1)^{2}}{3}}=\dfrac{6}{4(t^{2}+t+1)}=\dfrac{3}{2(t^{2}+t+1)}$.

Step 3: Match
Adding, $\dfrac{-(2t+1)+3}{2(t^{2}+t+1)}=\dfrac{1-t}{t^{2}+t+1}$, which is the integrand in t.
A minus sign or a logarithmic form in the arctangent part would not reproduce this.

Final Answer:
The arctangent form with coefficient $\sqrt{3}$ matches the integral. \[ \boxed{\text{(A) }-\dfrac{1}{2}\log |e^{2x}+e^{x}+1|+\sqrt{3}\tan ^{-1}\left(\dfrac{2e^{x}+1}{\sqrt{3}}\right)+c} \]