Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
Let $P(x)$ be a polynomial of least degree whose graph has three points of inflection $(-1,-1), (1,1)$ and a point with abscissa $0$ at which the curve is inclined to the axis of abscissa at an angle of $60°$. Then $\int_0^1 P(x)dx$ equals to:
$\frac{3\sqrt{3}+4}{14}$
$\frac{3\sqrt{3}}{7}$
$\frac{\sqrt{3}+\sqrt{7}}{14}$
$\frac{\sqrt{3}+2}{7}$
Step-by-Step Solution
Key Concept: Factoring out $x^{-n}$ and substituting $u = 1 + x^{-n}$ converts the integral into a standard power form.
Rewrite the integral as $\int x^{-2}(x^n + 1)^{-(n-1)/n}dx = \int x^{-2} \cdot x^{-n}(1 + x^{-n})^{-(n-1)/n}dx$. Substitute $u = 1 + x^{-n}$, so $du = -nx^{-n-1}dx$. The integral becomes $\frac{1}{n}\int u^{-(n-1)/n}du = \frac{1}{n} \cdot \frac{u^{1/n}}{1/n} + C = \frac{1}{1/n - 1}u^{1/n} + C = \frac{u^{1/n}}{(1-n)/n} + C$.
Correct Answer: 2