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: Understanding the Concept:
Over the symmetric limits \(-\sqrt{3}\) to \(\sqrt{3}\), the integral of an odd function is 0 and the integral of an even function is twice the integral from 0 to \(\sqrt{3}\).

Step 2: Split the numerator.
Odd terms: \(2x^9 - 5x^7 + 4x^3 - x\). Even terms: \(3x^8 + 9x^6 + 3\). The denominator \(x^2 + 3\) is even, so the odd terms give an odd integrand, whose integral is 0.

Step 3: Simplify the even part.
\[ \frac{3x^8 + 9x^6 + 3}{x^2 + 3} = \frac{3x^6(x^2 + 3) + 3}{x^2 + 3} = 3x^6 + \frac{3}{x^2 + 3} \]

Step 4: Integrate.
\[ I = 2\int_0^{\sqrt{3}}\left(3x^6 + \frac{3}{x^2 + 3}\right)dx = 2\left[\frac{3x^7}{7} + \sqrt{3}\tan^{-1}\frac{x}{\sqrt{3}}\right]_0^{\sqrt{3}} \]
Since \((\sqrt{3})^7 = 27\sqrt{3}\) and \(\tan^{-1}1 = \pi/4\):
\[ I = 2\left[\frac{81\sqrt{3}}{7} + \sqrt{3}\cdot\frac{\pi}{4}\right] = \frac{162\sqrt{3}}{7} + \sqrt{3}\,\frac{\pi}{2} \]

Step 5: Check the options.
Option (B) matches. Option (A) misses a factor of \(\sqrt{3}\) in the first term, and (C) and (D) have a wrong second term.

Final Answer:
The value of the integral is \(\dfrac{162\sqrt{3}}{7} + \sqrt{3}\dfrac{\pi}{2}\), option (B). \[ \boxed{\frac{162\sqrt{3}}{7} + \sqrt{3}\,\frac{\pi}{2}} \]