Not that difficult actually. Just a bit further of thinking.

Given the increasing occurrence of higher order questions in the past 10 years, I wouldn't be surprised if Cambridge threw out a similar one this year.

Hi Rei, here's an alternative method : (if you're not required to use i)'s result or if you were given this expression to integrate from the start)


∫ (e²ˣ / (e²ˣ + e⁻²ˣ)) dx

= ∫ (e²ˣ / (e²ˣ + e⁻²ˣ) × e²ˣ/e²ˣ) dx

= ∫ (e²ˣ(e²ˣ) / (e²ˣ(e²ˣ) + e²ˣ(e⁻²ˣ)) ) dx

= ∫ (e⁴ˣ/ (e⁴ˣ + 1)) dx

= ¼ ∫ (4e⁴ˣ/ (e⁴ˣ + 1)) dx

= ¼ ln (e⁴ˣ + 1) + c

Alternative presentation for the first method :


d/dx ln(e²ˣ + e⁻²ˣ) = (2e²ˣ - 2e⁻²ˣ))/(e²ˣ + e⁻²ˣ)

d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + 2

d/dx ln(e²ˣ + e⁻²ˣ) + 2 = (2e²ˣ - 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ) + (2e²ˣ + 2e⁻²ˣ)/(e²ˣ + e⁻²ˣ)

d/dx ln(e²ˣ + e⁻²ˣ) + 2 = 4e²ˣ/(e²ˣ + e⁻²ˣ)

∫ (d/dx ln(e²ˣ + e⁻²ˣ) + 2) dx = ∫4e²ˣ/(e²ˣ + e⁻²ˣ) dx

∫ (d/dx ln(e²ˣ + e⁻²ˣ)) dx + ∫ 2 dx = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx

ln(e²ˣ + e⁻²ˣ) + 2x = 4∫ e²ˣ/(e²ˣ + e⁻²ˣ) dx

¼(ln(e²ˣ + e⁻²ˣ) + 2x) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx


This can be simplified further into a single term.

¼(ln(e²ˣ + e⁻²ˣ) + 2x ln e) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx


¼(ln(e²ˣ + e⁻²ˣ) + ln e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx

¼ln((e²ˣ + e⁻²ˣ)e²ˣ) = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx

¼ln(e⁴ˣ + 1) + c = ∫e²ˣ/(e²ˣ + e⁻²ˣ) dx

(c is a arbitrary constant)