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

Answer 1

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

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

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

$=0+2{\int }_0^{\sqrt{2}}\frac{(3x^6-12x^2+1)}{x^2+2}.dx$

$[\because \text{If f(x) is an odd function then}{\int }_{-a}^{a}f(x)dx=0\text{\& if f(x) is and even function then}{\int }_{-a}^{a}f(x)dx=2{}^{}{\int }_0^{a}f(x).dx]$

$=2{\int }_0^{\sqrt{2}}\frac{3x^2(x^4-4)+1}{x^2+2}.dx$

$=2{\int }_0^{\sqrt{2}}\frac{3x^2(x^2+2)(x^2-2)}{x^2+2}.dx+2.{\int }_0^{\sqrt{2}}\frac{dx}{x^2+2}$

$=6{\int }_0^{\sqrt{2}}(x^4-2x^2).dx+2.{\int }_0^{\sqrt{2}}\frac{dx}{x^2+2}$

$=6{\int }_0^{\sqrt{2}}\left(x^4\right).dx-12{\int }_0^{\sqrt{2}}\left(x^2\right).dx+2.{\int }_0^{\sqrt{2}}\frac{dx}{x^2+(\sqrt{2}{)}^2}$

$=\frac{6}{5}[x^5{]}_0^{\sqrt{2}}-\frac{12}{3}[x^3{]}_0^{\sqrt{2}}+2.\frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]}_0^{\sqrt{2}}$

$=\frac{6}{5}(4\sqrt{2}-0)-4(2.\sqrt{2}-0)+\sqrt{2}\left[{\tan }^{-1}\right(1)-0]$

$=\frac{24}{5}\sqrt{2}-8\sqrt{2}+\sqrt{2}\times \frac{\pi }{4}$

$=\sqrt{2}(\frac{24}{5}-8)+\frac{\pi }{2\sqrt{2}}$

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