Question

${\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7+3x^6-10x^5-7x^3-12x^2+x+1}{x^2+2}dx=\cdots$

• A. $\frac{\pi }{2\sqrt{2}}+\frac{8\sqrt{2}}{5}$
• B. $\frac{\pi }{2\sqrt{2}}-\frac{16\sqrt{2}}{5}$
• C. $\frac{\pi }{\sqrt{2}}+\frac{16\sqrt{2}}{5}$
• D. $\frac{\pi }{\sqrt{2}}-\frac{8\sqrt{2}}{5}$

Answer 1

Answer: (B)

$\frac{\pi }{2\sqrt{2}}-\frac{16\sqrt{2}}{5}$

$I={\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7+3x^6-10x^5-7x^3-12x^2+x+1}{x^2+2}dx$
$I={\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7-10x^5-7x^3+x}{x^2+2}dx+{\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$
$I={\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$
$I=2{\int }_0^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$ ($\because {\int }_{-a}^{a}f\left(x\right)dx=\{\begin{array}{c}0\text{f(x) is odd}\\ 2{\int }_0^{a}f\left(x\right)dx\text{f((x) is even}\end{array}$)
$I=2{\int }_0^{\sqrt{2}}(3x^4-6x^2+\frac{1}{x^2+2})dx$
$=2[\frac{3x^5}{5}-2x^3+\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}{]}_0^{\sqrt{2}}$
$I=\frac{\pi }{2\sqrt{2}}-\frac{16\sqrt{2}}{5}$