Question Number 6959 by Tawakalitu. last updated on 03/Aug/16
$$\int\:{x}\:\left({x}\:+\:\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{dx}\: \\ $$
Answered by FilupSmith last updated on 03/Aug/16
$${u}={x}+\mathrm{2} \\ $$$${x}={u}−\mathrm{2} \\ $$$${du}={dx} \\ $$$$=\int\left({u}−\mathrm{2}\right){u}^{\mathrm{1}/\mathrm{4}} {du} \\ $$$$=\int\left({u}^{\mathrm{5}/\mathrm{4}} −\mathrm{2}{u}^{\mathrm{1}/\mathrm{4}} \right){du} \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}{u}^{\mathrm{9}/\mathrm{4}} −\mathrm{2}{u}^{\mathrm{5}/\mathrm{4}} +{c} \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\left({x}+\mathrm{2}\right)^{\mathrm{9}/\mathrm{4}} −\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{5}/\mathrm{4}} +{c} \\ $$