Question Detail

This integral can be solved using substitution. Let the denominator be u, so u = e2x - e-2x. Now, find the differential du: du = d⁄dx(e2x - e-2x) dx = (d⁄dx(e2x) - d⁄dx(e-2x)) dx = (2e2x - (-2e-2x)) dx = (2e2x + 2e-2x) dx = 2(e2x + e-2x) dx. From this, we can see that (e2x + e-2x) dx = 1⁄2 du. Substitute u and du into the integral: ∫ (1⁄u) * (1⁄2) du = 1⁄2 ∫ (1⁄u) du. The integral of 1⁄u is ln|u|. So, we have 1⁄2 ln|u| + C. Finally, substitute back u = e2x - e-2x: 1⁄2 ln|e2x - e-2x| + C. This matches option C.