Question Number 53 by surabhi last updated on 25/Jan/15
$$\mathrm{Evaluate}\:\int_{\mathrm{1}} ^{\mathrm{4}} \frac{\left({x}^{\mathrm{2}} +{x}\right)}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}{dx} \\ $$
Answered by surabhi last updated on 04/Nov/14
![[(x^2 +x)∙(√(2x+1))]_2 ^4 −∫_2 ^4 (2x+1)∙(√(2x+1))dx =(60−6(√5))−∫_2 ^4 (2x+1)^(3/2) dx =(60−6(√5))−(1/5)[(2x+1)^(5/2) ]_2 ^4 =(60−6(√5))−[((243)/5)−5(√5)] =((57)/5)−(√5)](https://www.tinkutara.com/question/Q54.png)
$$\left[\left({x}^{\mathrm{2}} +{x}\right)\centerdot\sqrt{\mathrm{2}{x}+\mathrm{1}}\right]_{\mathrm{2}} ^{\mathrm{4}} −\int_{\mathrm{2}} ^{\mathrm{4}} \left(\mathrm{2}{x}+\mathrm{1}\right)\centerdot\sqrt{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\left(\mathrm{60}−\mathrm{6}\sqrt{\mathrm{5}}\right)−\int_{\mathrm{2}} ^{\mathrm{4}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} {dx} \\ $$$$=\left(\mathrm{60}−\mathrm{6}\sqrt{\mathrm{5}}\right)−\frac{\mathrm{1}}{\mathrm{5}}\left[\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{5}/\mathrm{2}} \right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$=\left(\mathrm{60}−\mathrm{6}\sqrt{\mathrm{5}}\right)−\left[\frac{\mathrm{243}}{\mathrm{5}}−\mathrm{5}\sqrt{\mathrm{5}}\right] \\ $$$$=\frac{\mathrm{57}}{\mathrm{5}}−\sqrt{\mathrm{5}} \\ $$