Question Number 9107 by jainamanj98@gmail.com last updated on 18/Nov/16
Answered by FilupSmith last updated on 19/Nov/16
$$\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}+\mathrm{3}\sqrt{{x}}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{3}} {xdx}+\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{3}} {x}^{\mathrm{1}/\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{2}\left[{x}^{\mathrm{3}/\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{4}+\mathrm{2}\left(\sqrt{\mathrm{27}}−\mathrm{1}\right) \\ $$$$=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{27}} \\ $$$$=\mathrm{2}+\mathrm{2}\sqrt{\left(\mathrm{3}×\mathrm{9}\right)} \\ $$$$=\mathrm{2}+\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$\: \\ $$$$\left(\mathrm{2}\right) \\ $$$$\int\frac{{x}^{\mathrm{3}} }{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{dx} \\ $$$${u}={x}^{\mathrm{4}} +\mathrm{1} \\ $$$${du}=\mathrm{4}{x}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}}{{u}^{\mathrm{1}/\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int{u}^{−\mathrm{1}/\mathrm{3}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{2}}{u}^{\mathrm{2}/\mathrm{3}} \right)+{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} +{c} \\ $$