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
![∫_(√2) ^2 ((x^2 /(x+[x+1])))dx+∫_2 ^(√5) ((x^2 /(x+[x+1])))dx= ∫_(√2) ^2 ((x^2 /(x+2)))dx+∫_2 ^(√5) ((x^2 /(x+3)))dx= ∫_(√2) ^2 (((x^2 −4)/(x+2))+(4/(x+2)))dx+∫_2 ^(√5) (((x^2 −9)/(x+3))+(9/(x+3)))dx= ∫_(√2) ^2 (x−2)dx+∫_(√2) ^2 ((4/(x+2)))dx+∫_2 ^(√5) (x−3)dx+∫_2 ^(√5) ((9/(x+3)))dx [(x^2 /2)−2x]_(√2) ^2 +[4ln ∣x+2∣]_(√2) ^2 +[(x^2 /2)−3x]_2 ^(√5) +[9ln ∣x+3∣]_2 ^(√5) (2−4−1+2(√2)+4ln 4−4ln ((√2)+2))+((5/2)−3(√5)−2+6)+(9ln ((√5)+3)−9ln (5)) −3+2(√2)+4ln ((4/((√2)+2)))+((13)/2)−3(√5)+9ln ((((√5)+3)/5))](Q12922.png)
$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{3}}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}+\mathrm{2}}+\frac{\mathrm{4}}{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{9}}{{x}+\mathrm{3}}+\frac{\mathrm{9}}{{x}+\mathrm{3}}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left({x}−\mathrm{2}\right){dx}+\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{\mathrm{4}}{{x}+\mathrm{2}}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left({x}−\mathrm{3}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{9}}{{x}+\mathrm{3}}\right){dx} \\ $$$$\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} +\left[\mathrm{4ln}\:\mid{x}+\mathrm{2}\mid\right]_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} +\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}{x}\right]_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} +\left[\mathrm{9ln}\:\mid{x}+\mathrm{3}\mid\right]_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \\ $$$$\left(\mathrm{2}−\mathrm{4}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4ln}\:\mathrm{4}−\mathrm{4ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{2}\right)\right)+\left(\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{2}+\mathrm{6}\right)+\left(\mathrm{9ln}\:\left(\sqrt{\mathrm{5}}+\mathrm{3}\right)−\mathrm{9ln}\:\left(\mathrm{5}\right)\right) \\ $$$$−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4ln}\:\left(\frac{\mathrm{4}}{\sqrt{\mathrm{2}}+\mathrm{2}}\right)+\frac{\mathrm{13}}{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{9ln}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{5}}\right) \\ $$