Question

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

• A. $\frac{\pi }{2\sqrt{2}}+\frac{16\sqrt{2}}{5}$
• B. $\frac{\pi }{4\sqrt{2}}-\frac{8\sqrt{2}}{5}$
• C. $\frac{\pi }{4\sqrt{2}}+\frac{8\sqrt{2}}{5}$
• D. $\frac{\pi }{2\sqrt{2}}-\frac{16\sqrt{2}}{5}$

Answer 1

The correct option is D $\frac{\pi }{2\sqrt{2}}-\frac{16\sqrt{2}}{5}$

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