Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $k(x) = \int \frac{(x^2 + 1)dx}{\sqrt[4]{x^4 + 3x + 6}}$ and $k(-1) = \frac{1}{3\sqrt{2}}$, then the value of $k(-2)$ is
Step-by-Step Solution
Key Concept: Substitution $x^3 + 3x + 6 = t^3$ transforms the denominator to a linear term while the numerator matches the derivative structure.
For $k(x) = \int \frac{(x^2 + 1)dx}{(x^3 + 3x + 6)^{1/3}}$, use the substitution $x^3 + 3x + 6 = t^3$ so that $3(x^2 + 1)dx = 3t^2 dt$. Then $k(x) = \int \frac{t^2 dt}{t} = \frac{t^2}{2} + C = \frac{1}{2}(x^3 + 3x + 6)^{2/3} + C$. Using $k(-1) = \frac{1}{2}$ yields $C = 0$.
Correct Answer: 4