Question

The value of the integral \(\int _{-\sqrt{3}}^{\sqrt{3}}\frac{(2x^9+3x^8-5x^7+9x^6+4x^3-x+3)}{x^2+3}\,dx\) is

• A. \(\frac{162}{7}+\sqrt{3}\frac{π}{4}\)
• B. \(\frac{162\sqrt{3}}{7}+\sqrt{3}\frac{π}{2}\)
• C. \(\frac{162\sqrt{3}}{7}+\frac{π}{2}\)
• D. \(\frac{162\sqrt{3}}{7}-\frac{π}{2}\)

Hint:

Split into odd and even parts over the symmetric interval, so the odd part vanishes, then simplify the even part by division.

Answer 1

The Correct Option is B

Step 1: Use the property.
$\int_{-a}^{a} f(x)dx = \int_0^a [f(x) + f(-x)]dx$. For the numerator $N(x)$, $N(x) + N(-x) = 2(3x^8 + 9x^6 + 3)$, because odd terms cancel.

Step 2: Reduce.
Dividing by $x^2 + 3$ gives $2\left(3x^6 + \dfrac{3}{x^2 + 3}\right)$.

Step 3: Integrate from 0 to root 3.
$\int_0^{\sqrt3} 3x^6dx = \dfrac{3 \cdot 27\sqrt3}{7} = \dfrac{81\sqrt3}{7}$ and $\int_0^{\sqrt3}\dfrac{3\,dx}{x^2 + 3} = \sqrt{3}\cdot\dfrac{\pi}{4}$.

Step 4: Double.
\[ I = 2\left(\frac{81\sqrt3}{7} + \frac{\sqrt3\pi}{4}\right) = \frac{162\sqrt3}{7} + \frac{\sqrt3\pi}{2} \]

Final Answer:
Option (B). \[ \boxed{\frac{162\sqrt{3}}{7} + \sqrt{3}\,\frac{\pi}{2}} \]