Đáp án và lời giải
Đáp án:D
Lời giải:\begin{align*} I &= \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \dfrac{x^2+\left( 2x + \cos x \right) \cos x +1 - \sin x}{x + \cos x} \mathrm{\,d}x\\ &= \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \dfrac{(x+ \cos x)^2 + 1 - \sin x}{x + \cos x} \mathrm{\,d}x\\ &= \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \left( {x + \cos x} \right) \mathrm{\,d}x + \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \dfrac{1- \sin x}{x+ \cos x} \mathrm{\,d}x\\ &= \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \left( {x + \cos x} \right) \mathrm{\,d}x + \displaystyle \int \limits _0 ^{\tfrac{\pi}{2}} \dfrac{\mathrm{\,d}(x+ \cos x)}{x+ \cos x} \\ &= \left( \dfrac{x^2}{2} + \sin x \right) \bigg|_0^{\tfrac{\pi}{2}}+ \ln |x+ \cos x| \bigg|_0^{\tfrac{\pi}{2}}\\ &= \dfrac{\pi^2}{8}+1+\ln\dfrac{\pi}{2}\\ &=\dfrac{1}{8}\pi ^2 +1 - \ln \dfrac{2}{\pi} \end{align*} Suy ra $a=\dfrac{1}{8}$; $b=1$; $c=2$\\ Vậy $P=2.$
