4 ∫₀¹(1/(√(3+x²)+√(1+x²))) d x-3 log ₑ(√3) is equal to :

Solution

Maths sir · 5 steps

A sum of two square roots in the denominator is asking to be rationalised, so multiply by the conjugate first.

1. Rationalise. Multiply the top and bottom by √(3+x²)-√(1+x²). The denominator becomes (3+x²)-(1+x²) = 2, so the integrand is 1/2(√(3+x²)-√(1+x²)) and 4∫₀¹(…) dx = 2∫₀¹(√(3+x²)-√(1+x²))dx.

   ∫ √(a²+x²) dx = x/2√(a²+x²) + a²/2 ln(x+√(a²+x²))

2. First integral. With a² = 3 from 0 to 1: 1/2(2) + 3/2 ln 3 - 3/2 ln√3 = 1 + 3/4 ln 3.
3. Second integral. With a² = 1 from 0 to 1: √2/2 + 1/2 ln(1+√2) - 1/2 ln 1 = √2/2 + 1/2 ln(1+√2).
4. Combine. Twice the difference is 2 + 3/2 ln 3 - √2 - ln(1+√2). Now 3 ln√3 = 3/2 ln 3, so subtracting it removes the ln 3 term completely.
5. So. The expression equals 2-√2-ln(1+√2), which is option (a).

The idea to keep. When you see √A+√B in a denominator and A-B is a constant, rationalise. It turns an ugly fraction into standard √(a²+x²) integrals.

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