Let I(x)=∫ (x+1)/x(1+xeˣ)²dx,x>0. If limI(x)=0, then I(1) is equal to :

Solution

Maths sir · 4 steps

The (x+1) on top is the hint: put t = xeˣ and everything turns into partial fractions in t.

1. Choose the substitution. Let t = xeˣ. Then dt = eˣ(x+1) dx, so (x+1) dx = dt/eˣ. The leftover x in the denominator pairs with this eˣ to make xeˣ = t.

   I = ∫ dt/t(1+t)² = ∫ (1/t - 1/(1+t) - 1/(1+t)²) dt

2. Integrate. Integrating term by term gives I = ln t/(1+t) + 1/(1+t) + C, with t = xeˣ.
3. Fix the constant. As x → ∞, t → ∞. Then t/(1+t) → 1, so the log tends to 0, and 1/(1+t) → 0. The limit is C, so C = 0.
4. So. At x = 1, t = e, and I(1) = ln e/(e+1) + 1/(e+1) = 1 + 1/(e+1) - ln(e+1) = (e+2)/(e+1) - logₑ(e+1), which is option (c).

The idea to keep. When the numerator is (x+1) and xeˣ appears, set t = xeˣ because its derivative is eˣ(x+1). A limit condition on I(x) is there to fix the constant C.

Answer from the NTA final answer key. Solution by PrepWiser.