Question

${\int }_{-2}^2\frac{3x^5+4x^3+2x^2+x+20}{x^2+4}dx=?$

• A. $8+3\pi$
• B. $3+8\pi$
• C. $4+3\pi$
• D. $3+4\pi$

Answer 1

The correct option is A $8+3\pi$

$I=0+2{\int }_0^2\frac{2x^2+20}{x^2+4}dx,$ by above prop.
$=2.2{\int }_0^2\frac{x^2+4+6}{x^2+4}dx=4({\int }_0^{2}dx+6{\int }_0^2\frac{dx}{x^2+4})$
$=4{[x+6.\frac{1}{2}{\tan }^{-1}\frac{x}{2}]}_0^2=4[2+3.\frac{\pi }{4}]=8+3\pi$