|
|
Ảo thuật một chút nhé :)
Ta có
$ \frac{x^2}{\sqrt{4+2x}}$
$=\frac{1}{15}. \frac{(3x^2-8x+32)+(12x^2+8x-32)}{\sqrt{4+2x}}$
$=\frac{1}{15}. \frac{(3x^2-8x+32)+(2x+4)(6x-8)}{\sqrt{4+2x}}$
$=\frac{1}{15}. \frac{1}{\sqrt{4+2x}}(3x^2-8x+32)+\frac{1}{15}.\sqrt{4+2x}(6x-8)$
$=\frac{1}{15}.\left (\sqrt{4+2x} \right )'(3x^2-8x+32)+\frac{1}{15}.\sqrt{4+2x}(3x^2-8x+32)'$
$=\frac{1}{15}.\left (\sqrt{4+2x}(3x^2-8x+32) \right )'$
Vậy
$I=\frac{1}{15}\int\limits_{0}^{1}\left (\sqrt{4+2x}(3x^2-8x+32) \right )'dx$
$I=\frac{1}{15}\sqrt{4+2x}(3x^2-8x+32)|_0^1 $
$\boxed{I=\displaystyle \frac{27\sqrt 6 - 64}{15}}$
|