JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
∫0^∞ 6/(e³ˣ+6 e²ˣ+11 eˣ+6) d x=
- (a) log ₑ(256/81)
- (b) log ₑ(64/27)
- (c) log ₑ(32/27)
- (d) log ₑ(512/81)
Answer and solution
Answer (c) log ₑ(32/27)
Solution
Maths sir · 4 steps
Put t = eˣ; the cubic splits into (t+1)(t+2)(t+3) and partial fractions finish it.
- Substitute. Let t = eˣ, so dx = dt/t. When x = 0, t = 1, and as x → ∞, t → ∞. The denominator t³ + 6t² + 11t + 6 factors as (t+1)(t+2)(t+3).
∫₁^∞ (6 dt)/t(t+1)(t+2)(t+3) = ∫₁^∞(1/t - 3/(t+1) + 3/(t+2) - 1/(t+3))dt
- Partial fractions. Use the cover-up rule. At t = 0 the coefficient is 6/(1 × 2 × 3) = 1. At t = -1 it is 6/(-1)(1)(2) = -3. At t = -2 it is 6/(-2)(-1)(1) = 3. At t = -3 it is 6/(-3)(-2)(-1) = -1.
- Integrate. The antiderivative is ln t(t+2)³/(t+1)³(t+3). As t → ∞ the fraction tends to 1, so the upper limit gives ln 1 = 0. At t = 1 the fraction is (1 × 27)/(8 × 4) = 27/32.
- So. The integral is 0 - ln 27/32 = ln 32/27, which is option (c).
The idea to keep. When the integrand is built from powers of eˣ, substitute t = eˣ and remember the extra 1/t from dx. Combine the logs into one fraction before taking the limit at infinity.
Answer from the NTA final answer key. Solution by PrepWiser.