Question Number 75081 by Rio Michael last updated on 07/Dec/19
$${Evaluate}\: \\ $$$$\:\int_{\mathrm{1}} ^{\mathrm{3}\:} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}}\:{dx} \\ $$
Answered by mr W last updated on 07/Dec/19
![∫_1 ^(3 ) (x^2 /(1+x)) dx =∫_1 ^(3 ) ((x^2 −1+1)/(1+x)) dx =∫_1 ^(3 ) (x−1+(1/(1+x))) dx =[(x^2 /2)−x+ln (x+1)]_1 ^3 =((3^2 −1^2 )/2)−3+1+ln (3+1)−ln 2 =2+ln 2](https://www.tinkutara.com/question/Q75085.png)
$$\:\int_{\mathrm{1}} ^{\mathrm{3}\:} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}}\:{dx} \\ $$$$\:=\int_{\mathrm{1}} ^{\mathrm{3}\:} \frac{{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\mathrm{1}+{x}}\:{dx} \\ $$$$\:=\int_{\mathrm{1}} ^{\mathrm{3}\:} \left({x}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\:{dx} \\ $$$$\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$\:=\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}+\mathrm{1}+\mathrm{ln}\:\left(\mathrm{3}+\mathrm{1}\right)−\mathrm{ln}\:\mathrm{2} \\ $$$$=\mathrm{2}+\mathrm{ln}\:\mathrm{2} \\ $$
Commented by Rio Michael last updated on 07/Dec/19
$${thank}\:{you}\:{sir} \\ $$