JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
The integral ∫ 1/∜(x-1)³(x+2)⁵ dx is equal to : (where C is a constant of integration)
- (a) 3/4((x+2)/(x-1))^(1/4)+C
- (b) 3/4((x+2)/(x-1))^(5/4)+C
- (c) 4/3((x-1)/(x+2))^(1/4)+C
- (d) 4/3((x-1)/(x+2))^(5/4)+C
Answer and solution
Answer (c) 4/3((x-1)/(x+2))^(1/4)+C
Solution
Maths sir · 4 steps
The powers add up to 2, so pull out (x+2)² and let the ratio (x-1)/(x+2) be your t.
- Rewrite the integrand. The integrand is (x-1)^(-3/4)(x+2)^(-5/4). The two powers add to -2, so write it as 1/(x+2)²((x-1)/(x+2))^(-3/4).
- Choose the substitution. Let t = (x-1)/(x+2). Then dt/dx = ((x+2)-(x-1))/(x+2)² = 3/(x+2)², so dx/(x+2)² = dt/3.
∫ 1/(x+2)²((x-1)/(x+2))^(-3/4)dx = 1/3∫ t^(-3/4) dt = 1/3·t^(1/4)/(1/4) + C
- Simplify. 1/3× 4 t^(1/4) = 4/3t^(1/4). Put back t = (x-1)/(x+2).
- So. The integral is 4/3((x-1)/(x+2))^(1/4) + C, which is option (c).
The idea to keep. For (x-a)ᵐ(x-b)ⁿ with m + n = -2, divide inside and out by (x-b)² and substitute t = (x-a)/(x-b). Its derivative brings exactly that square.
Answer from the NTA final answer key. Solution by PrepWiser.