The integral ∫ [(x/2)ˣ+(2/x)ˣ]ln (ex/2)dx is equal to :

Solution

Maths sir · 4 steps

Notice that ln ex/2 is exactly the derivative of x ln x/2, the exponent of both terms.

1. Spot the exponent. Write (x/2)ˣ = eᵗ with t = x ln x/2. Then (2/x)ˣ = e⁻ᵗ, because it is the reciprocal.
2. Differentiate t. By the product rule, dt/dx = ln x/2 + x × 1/x = ln x/2 + ln e = ln ex/2. That is exactly the factor sitting outside the bracket.

   ∫ (eᵗ + e⁻ᵗ) dt = eᵗ - e⁻ᵗ + C

3. Put x back. Replacing eᵗ by (x/2)ˣ and e⁻ᵗ by (2/x)ˣ gives (x/2)ˣ - (2/x)ˣ + C.
4. So. The answer is (b). You can check it: differentiating (2/x)ˣ = e⁻ᵗ gives -(2/x)ˣ ln ex/2, so the minus sign turns it into +.

The idea to keep. For a power like f(x)ᵍ⁽ˣ⁾, write it as eᵍˡⁿᶠ and look for the derivative of g ln f in the integrand. If it is there, substitute and the integral becomes routine.

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