Question Number 36541 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$$\int\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} {dx} \\ $$
Commented by prof Abdo imad last updated on 03/Jun/18
$${let}\:{put}\:\:{A}_{{n}} =\:\int\:\left({x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} \:{dx}\:{changement} \\ $$$${x}\:={sh}\left({t}\right)\:{give}\: \\ $$$${A}_{{n}} =\:\int\:\left({sht}\:\:+{cht}\right)^{{n}} \:{ch}\:{dt} \\ $$$$=\int\:\left\{\:\frac{{e}^{{t}} \:−{e}^{−{t}} \:+\:{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right\}^{{n}} \:{cht}\:{dt} \\ $$$$=\int\:\:{e}^{{nt}} \:\left(\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:{e}^{\left({n}+\mathrm{1}\right){t}} {dt}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\:{e}^{\left({n}−\mathrm{1}\right){t}} {dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}{e}^{\left({n}+\mathrm{1}\right){t}} \:\:\:+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}{e}^{\left({n}−\mathrm{1}\right){t}} \:\:+{c} \\ $$$${but}\:{e}^{\left({n}+\mathrm{1}\right){t}} \:={e}^{\left({n}+\mathrm{1}\right){ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=\left({x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}+\mathrm{1}} \:{and}\:{by}?{tha}\:{same} \\ $$$${e}^{\left({n}−\mathrm{1}\right){t}} \:=\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}−\mathrm{1}} \:\:\Rightarrow \\ $$$${A}_{{n}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left\{\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}+\mathrm{1}} \\ $$$$+\:\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\left\{\:{x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}−\mathrm{1}} \:\:+{c}\:\:\:\:{if}\:{n}\neq\mathrm{1} \\ $$