Question Number 12908 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17
Answered by 433 last updated on 07/May/17
$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{3}}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}+\mathrm{2}}+\frac{\mathrm{4}}{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{9}}{{x}+\mathrm{3}}+\frac{\mathrm{9}}{{x}+\mathrm{3}}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left({x}−\mathrm{2}\right){dx}+\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{\mathrm{4}}{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left({x}−\mathrm{3}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{9}}{{x}+\mathrm{3}}\right){dx} \\ $$$$\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} +\left[\mathrm{4ln}\:\mid{x}+\mathrm{2}\mid\right]_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} +\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}{x}\right]_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} +\left[\mathrm{9ln}\:\mid{x}+\mathrm{3}\mid\right]_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \\ $$$$\left(\mathrm{2}−\mathrm{4}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4ln}\:\mathrm{4}−\mathrm{4ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{2}\right)\right)+\left(\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{2}+\mathrm{6}\right)+\left(\mathrm{9ln}\:\left(\sqrt{\mathrm{5}}+\mathrm{3}\right)−\mathrm{9ln}\:\left(\mathrm{5}\right)\right) \\ $$$$−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4ln}\:\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}+\mathrm{2}}\right)+\frac{\mathrm{13}}{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{9ln}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{5}}\right) \\ $$