JEE Main 2020 · 9 January · morning · Single correct · +4 / −1

The integral ∫ dx/(x+4)^(8/7)(x-3)^(6/7) is equal to :
(where C is a constant of integration)

  1. (a) 1/2((x-3)/(x+4))^(3/7)+C
  2. (b) ((x-3)/(x+4))^(1/7)+C
  3. (c) -1/13((x-3)/(x+4))^(13/7)+C
  4. (d) -((x-3)/(x+4))^(-1/7)+C
Answer and solution

Answer (b) ((x-3)/(x+4))^(1/7)+C

Solution

Maths sir · 4 steps

Powers add up to 2, so pull out (x+4)² and put t = (x-3)/(x+4).

  1. Spot the pattern. The two powers add up to 8/7 + 6/7 = 2. So the denominator can be written as (x+4)² times a power of the ratio (x-3)/(x+4).
  2. Rewrite the denominator. (x+4)^(8/7)(x-3)^(6/7) = (x+4)²((x-3)/(x+4))^(6/7), since (x+4)^(8/7) = (x+4)² × (x+4)^(-6/7).
  3. Substitute. Let t = (x-3)/(x+4). By the quotient rule, dt/dx = ((x+4)-(x-3))/(x+4)² = 7/(x+4)², so dx/(x+4)² = dt/7.

    ∫ dx/((x+4)² t^(6/7)) = 1/7∫ t^(-6/7) dt = 1/7·t^(1/7)/(1/7) + C = t^(1/7) + C

  4. So. The integral is ((x-3)/(x+4))^(1/7) + C, which is option (b).

The idea to keep. For ∫ dx/((x+a)ᵐ (x+b)ⁿ) with m + n = 2, take out (x+a)² and substitute t = (x+b)/(x+a). Its derivative brings back exactly 1/(x+a)².

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

Study Indefinite Integration

Asked about 60 times in JEE Main since 2020. This question is from 2020.