Question

Evaluate \( \int \frac{e^{2x} - 1}{e^{2x} + 1} \, dx \).

Hint:

For integrals involving exponentials in the numerator and denominator, try substitution and algebraic manipulation to simplify the fraction.

Answer 1

Step 1: Substituting \( t = e^{2x} \). Let: \[ t = e^{2x} \Rightarrow dt = 2e^{2x} dx = 2t \, dx. \] Rewriting the integral: \[ I = \int \frac{e^{2x} - 1}{e^{2x} + 1} \, dx. \] Step 2: Splitting the fraction. Rewrite the numerator: \[ e^{2x} - 1 = (e^{2x} + 1) - 2. \] Thus, the integral becomes: \[ I = \int \frac{(e^{2x} + 1) - 2}{e^{2x} + 1} \, dx. \] Splitting: \[ I = \int \left( 1 - \frac{2}{e^{2x} + 1} \right) dx. \] Step 3: Solving the integral. The first term: \[ \int 1 \, dx = x. \] For the second term, let: \[ t = e^{2x} + 1 \Rightarrow dt = 2e^{2x} dx = 2(t - 1) dx. \] Rearrange: \[ dx = \frac{dt}{2(t - 1)}. \] Thus: \[ \int \frac{2}{e^{2x} + 1} \, dx = \int \frac{2}{t} \cdot \frac{dt}{2} = \int \frac{dt}{t} = \log |t|. \] Substituting back \( t = e^{2x} + 1 \): \[ \log |e^{2x} + 1|. \] Step 4: Conclusion. \[ I = \log (e^{2x} + 1) - x + C. \] Thus, the correct answer is \( \mathbf{\log (e^{2x} + 1) - x + C} \).