Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $\int (x^9 + x^6 + x^3)(2x^6 + 3x^3 + 6)^{1/3} dx = a(2x^9 + 3x^6 + 6x^3)^{4/3} + c$, then the value of $48a$ must be
Step-by-Step Solution
Key Concept: Use substitution $t = 2x^6 + 3x^6 + 6x^3$ where the differential matches the remaining factor in the integrand.
Let $t = 2x^6 + 3x^6 + 6x^3$. Then $dt = 18(x^5 + x^5 + x^2)dx = 18(x^5 + x^5 + x^2)dx$. The integral becomes $I = \int (x^8 + x^6 + x^2)^{1/3}(2x^6 + 3x^6 + 6x)^{1/3}dx = \frac{1}{18}\int t^{1/3}dt = \frac{1}{18} \cdot \frac{3}{4}t^{4/3} = \frac{1}{24}t^{4/3} + c$.
Correct Answer: 2,3