If ∫ 1/⁵√(x-1)⁴(x+3)⁶ dx=A((α x-1)/(β x+3))^B+C, where C is the constant of integration, then the value of α+β+20AB is _________.

Solution

Maths sir · 4 steps

The powers add up to 2, so pull out (x+3)² and substitute t = (x-1)/(x+3).

1. Rewrite the integrand. The integrand is (x-1)^(-4/5)(x+3)^(-6/5). The two powers add up to -2, so split (x+3)^(-6/5) = (x+3)^(4/5)(x+3)⁻². This gives ((x-1)/(x+3))^(-4/5)1/(x+3)².
2. Substitute. Let t = (x-1)/(x+3). Then dt/dx = ((x+3)-(x-1))/(x+3)² = 4/(x+3)², so dx/(x+3)² = dt/4.

   ∫ 1/4 t^(-4/5) dt = 1/4 × 5 t^(1/5) + C = 5/4((x-1)/(x+3))^(1/5) + C

3. Match the form. Comparing with A((α x-1)/(β x+3))^B gives α = 1, β = 1, A = 5/4 and B = 1/5.
4. So. α + β + 20AB = 1 + 1 + 20 × 5/4 × 1/5 = 2 + 5 = 7.

The common slip: 22. Forgetting the factor 1/4 from dt gives A = 5, so 20AB = 20 and the total becomes 22, not 7.

The idea to keep. When the integrand is (x-a)ᵐ(x-b)ⁿ with m + n = -2, take (x-b)² out and put t = (x-a)/(x-b). The integral becomes a single power of t.

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