Question Number 57241 by Aditya789 last updated on 01/Apr/19
$$\int\frac{×\sqrt{\mathrm{x}+\mathrm{1}}}{\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19
$${x}+\mathrm{1}={t}^{\mathrm{2}} \rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}×\mathrm{2}{tdt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int{t}^{\mathrm{2}} {dt}−\mathrm{4}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int{t}^{\mathrm{2}} {dt}−\mathrm{4}\int{dt}+\mathrm{4}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{2}×\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{4}{t}+\mathrm{4}{tan}^{−\mathrm{1}} \left({t}\right)+{c} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{4}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{4}{tan}^{−\mathrm{1}} \left(\sqrt{{x}+\mathrm{1}}\right)+{c} \\ $$$$ \\ $$