Question

The value of ${\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7+3x^6-10x^5-7x^3-12x^2+x+1}{x^2+2}dx$ is

• A. $0$
• B. $\frac{-16}{5}\sqrt{2}+\frac{\pi }{2\sqrt{2}}$
• C. $2{\int }_0^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$
• D. $\frac{-14}{5}\sqrt{2}+\frac{\pi }{2\sqrt{2}}$

Answer 1

Answer: (B), (C)

$\frac{-16}{5}\sqrt{2}+\frac{\pi }{2\sqrt{2}}$
$2{\int }_0^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$

$I={\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7+3x^6-10x^5-7x^3-12x^2+x+1}{x^2+2}dx$
$={\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^7-10x^5-7x^3+x}{x^2+2}dx+{\int }_{-\sqrt{2}}^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx$
$=0+2{\int }_0^{\sqrt{2}}\frac{3x^6-12x^2+1}{x^2+2}dx=2{\int }_0^{\sqrt{2}}\frac{3x^2(x^4-4)+1}{x^2+2}dx$
$=2{\int }_0^{\sqrt{2}}(3x^2(x^2-2)+\frac{1}{x^2+2})dx=2{\int }_0^{\sqrt{2}}(3x^4-6x^2+\frac{1}{x^2+2})dx$
$=2{(\frac{3x^5}{5}-2x^3+\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x}{2}\right))}_0^{\sqrt{2}}=\frac{-16}{5}\sqrt{2}+\frac{\pi }{2\sqrt{2}}$