MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Use symmetry of definite integrals over $[-a,a]$ to eliminate odd integrands, then integrate the remaining even part.Explanation:The given integral is$$I = \int_{-\sqrt{3}}^{\sqrt{3}} \frac{2x^9+3x^8-5x^7+9x^6+4x^3-x+3}{x^2+3}\,dx.$$Since $x^2+3$ is even, terms with odd powers of $x$ in the numerator become odd functions after division.The odd terms are $2x^9$, $-5x^7$, $4x^3$, and $-x$, and their integral over $[-\sqrt{3},\sqrt{3}]$ is $0$.So only the even part remains:$$I = \int_{-\sqrt{3}}^{\sqrt{3}} \frac{3x^8+9x^6+3}{x^2+3}\,dx.$$Now simplify the integrand.$$3x^8+9x^6+3 = 3x^6(x^2+3)+3,$$so$$\frac{3x^8+9x^6+3}{x^2+3} = 3x^6 + \frac{3}{x^2+3}.$$Thus,$$I = \int_{-\sqrt{3}}^{\sqrt{3}} 3x^6\,dx + \int_{-\sqrt{3}}^{\sqrt{3}} \frac{3}{x^2+3}\,dx.$$First,$$\int_{-\sqrt{3}}^{\sqrt{3}} 3x^6\,dx = 3\left[\frac{x^7}{7}\right]_{-\sqrt{3}}^{\sqrt{3}} = \frac{3}{7}\left[(\sqrt{3})^7 - (-\sqrt{3})^7\right].$$Since $(\sqrt{3})^7 = 27\sqrt{3}$, this becomes$$\frac{3}{7}(54\sqrt{3}) = \frac{162\sqrt{3}}{7}.$$Next,$$\int_{-\sqrt{3}}^{\sqrt{3}} \frac{3}{x^2+3}\,dx = \left[\sqrt{3}\tan^{-1}\!\left(\frac{x}{\sqrt{3}}\right)\right]_{-\sqrt{3}}^{\sqrt{3}}.$$Evaluating,$$\sqrt{3}\left(\tan^{-1}(1)-\tan^{-1}(-1)\right) = \sqrt{3}\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = \frac{\sqrt{3}\pi}{2}.$$Therefore,$$I = \frac{162\sqrt{3}}{7} + \frac{\sqrt{3}\pi}{2}.$$Answer:The correct option is B: $\frac{162\sqrt{3}}{7} + \frac{\sqrt{3}\pi}{2}$.