Question

Evaluate: \[ \int \frac{e^{2x}-1}{e^{2x}+e^x+1}\,dx \]

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

Hint:

For integrals containing \(e^x\), always try the substitution \(e^x=t\). It converts exponential expressions into algebraic fractions.

Answer 1

The Correct Option is B

Concept: For integrals involving exponentials, substitution simplifies the expression considerably.

Step 1: Use substitution \(e^x=t\). Let \[ e^x=t \] Then, \[ e^{2x}=t^2 \] and \[ dx=\frac{dt}{t} \] Thus, \[ I=\int \frac{t^2-1}{t^2+t+1}\cdot \frac{dt}{t} \] Factor numerator: \[ t^2-1=(t-1)(t+1) \] So, \[ I=\int \frac{(t-1)(t+1)}{t(t^2+t+1)}dt \] Using partial fractions, \[ \frac{t^2-1}{t(t^2+t+1)} = -\frac1t+\frac{2t+1}{t^2+t+1} \] Hence, \[ I=\int -\frac1t\,dt+\int \frac{2t+1}{t^2+t+1}\,dt \]

Step 2: Integrate each term. \[ \int -\frac1t\,dt=-\log|t| \] and \[ \int \frac{2t+1}{t^2+t+1}dt = \log(t^2+t+1) \] Therefore, \[ I= \log(t^2+t+1)-\log t+c \] Substituting \(t=e^x\), \[ I= \log(e^{2x}+e^x+1)-x+c \] Hence the answer is \[ \boxed{ \log(e^{2x}+e^x+1)-x+c } \]