Question

10．求定积分${∫}_{-1}^{0}$$\frac{{x}^{2}}{{x}^{2}+2x}$dx的值．

Answer 1

分析 变形可得原式=${∫}_{-1}^{0}$（1-$\frac{2x}{{x}^{2}+2x}$）dx=${∫}_{-1}^{0}$（1-$\frac{2}{x+2}$）dx=${∫}_{-1}^{0}$dx-${∫}_{-1}^{0}$$\frac{2}{x+2}$dx=[x-2ln（x+2）]${|}_{-1}^{0}$，代值计算可得．

解答 解：${∫}_{-1}^{0}$$\frac{{x}^{2}}{{x}^{2}+2x}$dx=${∫}_{-1}^{0}$$\frac{{x}^{2}+2x-2x}{{x}^{2}+2x}$dx
=${∫}_{-1}^{0}$（1-$\frac{2x}{{x}^{2}+2x}$）dx=${∫}_{-1}^{0}$（1-$\frac{2}{x+2}$）dx
=${∫}_{-1}^{0}$dx-${∫}_{-1}^{0}$$\frac{2}{x+2}$dx=[x-2ln（x+2）]${|}_{-1}^{0}$
=（0-2ln2）-（-1-2ln1）=1-2ln2

点评 本题考查定积分的求解，属基础题．