If
where, is a constant of integration, then the function is equal to
where, is a constant of integration, then the function is equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
We need to find f(x) such that ∫ dx/[x³(1+x⁶)^(2/3)] = x·f(x)·(1+x⁶)^(1/3) + C.
Differentiate the right side and set it equal to the integrand. Factor out (1+x⁶)^(1/3) and note the integrand can be rewritten by factoring x⁶ from (1+x⁶)^(2/3):
1/[x³(1+x⁶)^(2/3)] = 1/[x³ · x⁴(1/x⁶ + 1)^(2/3)] = 1/[x⁷(1+x⁻⁶)^(2/3)]
Better approach: write (1+x⁶)^(2/3) = x⁴(x⁻⁶+1)^(2/3), so the integrand = x⁻⁷(1+x⁻⁶)^(-2/3).
Let t = 1 + x⁻⁶, then dt = -6x⁻⁷ dx, so x⁻⁷ dx = -dt/6.
Integral = ∫ t^(-2/3)·(-dt/6) = (-1/6)·3t^(1/3) + C = (-1/2)(1+x⁻⁶)^(1/3) + C.
Rewrite: (-1/2)(1+x⁻⁶)^
Differentiate the right side and set it equal to the integrand. Factor out (1+x⁶)^(1/3) and note the integrand can be rewritten by factoring x⁶ from (1+x⁶)^(2/3):
1/[x³(1+x⁶)^(2/3)] = 1/[x³ · x⁴(1/x⁶ + 1)^(2/3)] = 1/[x⁷(1+x⁻⁶)^(2/3)]
Better approach: write (1+x⁶)^(2/3) = x⁴(x⁻⁶+1)^(2/3), so the integrand = x⁻⁷(1+x⁻⁶)^(-2/3).
Let t = 1 + x⁻⁶, then dt = -6x⁻⁷ dx, so x⁻⁷ dx = -dt/6.
Integral = ∫ t^(-2/3)·(-dt/6) = (-1/6)·3t^(1/3) + C = (-1/2)(1+x⁻⁶)^(1/3) + C.
Rewrite: (-1/2)(1+x⁻⁶)^
[JEE Main 2019 · 8 April · Shift II]