JEE Main 2024 · 27 January · evening · Single correct · +4 / −1
The integral∫ (x⁸-x²)dx/(x¹²+3x⁶+1)tan⁻¹(x³+1/x³) is equal to:
- (a) logₑ(|tan⁻¹(x³+1/x³)|)^(1/3)+C
- (b) logₑ(|tan⁻¹(x³+1/x³)|)+C
- (c) logₑ(|tan⁻¹(x³+1/x³)|)^(1/2)+C
- (d) logₑ(|tan⁻¹(x³+1/x³)|)³+C
Answer and solution
Answer (a) logₑ(|tan⁻¹(x³+1/x³)|)^(1/3)+C
Solution
Maths sir · 4 steps
Substitute the inverse tan in the denominator: its derivative is sitting right there in the numerator.
- Choose the substitution. Put t = tan⁻¹(x³ + 1/x³). Differentiate with the chain rule: dt/dx = 1/(1 + (x³ + 1/x³)²) × 3(x² - 1/x⁴).
- Clean up both factors. The first factor: 1 + x⁶ + 2 + 1/x⁶ = (x¹² + 3x⁶ + 1)/x⁶. The second: x² - 1/x⁴ = (x⁶ - 1)/x⁴.
dt = 3x⁶/(x¹²+3x⁶+1) × (x⁶-1)/x⁴ dx = 3(x⁸ - x²)/(x¹²+3x⁶+1) dx
- Rewrite the integral. So ((x⁸ - x²) dx)/(x¹²+3x⁶+1) = dt/3, and the integral becomes 1/3∫ dt/t = 1/3 ln|t| + C.
- So. 1/3 ln|t| = ln|t|^(1/3), giving logₑ(|tan⁻¹(x³ + 1/x³)|)^(1/3) + C, which is option (a).
The idea to keep. When a messy function sits in the denominator, differentiate it. If its derivative matches the rest of the integrand, you get ∫ dt/t. Track the constant the chain rule leaves behind.
Answer from the NTA final answer key. Solution by PrepWiser.