If ∫₀¹ 1/(√(3+x)+√(1+x)) d x=a+b √2+c √3, where a, b, c are rational numbers, then 2 a+3 b-4 c is equal to :

Solution

Maths sir · 4 steps

Rationalise first: the two quantities under the roots differ by exactly 2, so the denominator vanishes.

1. Rationalise. Multiply top and bottom by √(3+x)-√(1+x). The denominator becomes (3+x)-(1+x) = 2, so the integrand is 1/2(√(3+x)-√(1+x)).

   ∫₀¹ (√(3+x)-√(1+x))/2 dx = 1/3[(3+x)^(3/2)-(1+x)^(3/2)]₀¹

2. Put in limits. At x=1: 4^(3/2)-2^(3/2) = 8-2√2. At x=0: 3^(3/2)-1 = 3√3-1. Subtracting gives 9-2√2-3√3, and dividing by 3 gives 3-2/3√2-√3.
3. Match coefficients. Comparing with a+b√2+c√3 gives a=3, b=-2/3, c=-1.
4. So. 2a+3b-4c = 6-2+4 = 8, which is option (d).

The idea to keep. When the denominator is a sum of two square roots whose insides differ by a constant, rationalise. The integral turns into plain powers of linear terms.

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