Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
The value of $\int_0^1 \frac{2x^2 + 3x + 3}{(x+1)(x^2 + 2x + 2)} dx$ is:
$\frac{\pi}{4} + 2\log 2 - \tan^{-1} 2$
$\frac{\pi}{4} + 2\log 2 - \tan^{-1} \frac{1}{3}$
$2\log 2 - \cot^{-1} 3$
$\frac{\pi}{4} + \log 4 + \cot^{-1} 2$
Step-by-Step Solution
Key Concept: Partial fraction decomposition combined with standard integral formulas reduces complex rational integrals to logarithmic and inverse trigonometric forms.
The integral $\int_0^1 \frac{2x^2 + 3x + 3}{(x+1)(x^2+2x+2)}\,dx$ is split into partial fractions: $\int_0^1 \left(\frac{2}{x+1} - \frac{1}{x^2+2x+2}\right)dx = [2\ln(x+1) - \tan^{-1}(x+1)]_0^1 = 2\ln 2 - \tan^{-1}2 - \tan^{-1}1 = 2\ln 2 - \cot^{-1}3$.
Correct Answer: 1,4