# EvAluate-x-2-9-9-dx-

Question Number 11616 by Nayon last updated on 29/Mar/17
$${EvAluate}\:\int\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{9}} {dx} \\$$
Answered by Joel576 last updated on 29/Mar/17
$$\mathrm{Let}\:{u}\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{9} \\$$$$\frac{{du}}{{dx}}\:=\:\mathrm{2}{x}\:\:\Leftrightarrow\:\:{dx}\:=\:\frac{{du}}{\mathrm{2}{x}}\:\: \\$$$$\\$$$$\int\:{u}^{\mathrm{9}} \:{dx} \\$$$$=\:\int\:\frac{{u}^{\mathrm{9}} }{\mathrm{2}{x}}\:{du} \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:.\:\int\:{u}^{\mathrm{9}} \:{du} \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:.\:\frac{\mathrm{1}}{\mathrm{10}}{u}^{\mathrm{10}} \\$$$$=\:\frac{\left({x}^{\mathrm{2}} \:+\:\mathrm{9}\right)^{\mathrm{10}} }{\mathrm{20}{x}}\:+\:{C} \\$$
Commented by Nayon last updated on 29/Mar/17
$${Biggg}\:{wroooong}\:{joel}!! \\$$$$\\$$
Commented by Nayon last updated on 29/Mar/17
$${u}={x}^{\mathrm{2}} +\mathrm{9} \\$$$$=>{x}^{\mathrm{2}} ={u}−\mathrm{9} \\$$$$=>{x}=\pm\sqrt{{u}−\mathrm{9}} \\$$$$=>\mathrm{2}{x}=\pm\mathrm{2}\sqrt{{u}−\mathrm{9}} \\$$$$\\$$
Commented by linkelly0615 last updated on 29/Mar/17
$$… \\$$$${uh}… \\$$$$\frac{{d}}{{dx}}\left[\frac{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{10}} }{\mathrm{20}{x}}\right]\neq\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{9}} \\$$
Commented by Nayon last updated on 29/Mar/17
$${I}\:{need}\:{it}'{s}\:{ans}.\:{urgently}\:{pls}\:{help} \\$$$${me}…… \\$$$$\\$$
$${I}\:{thik}\:{it}\:{does}\:{not}\:{have}\:{a}\:{simplied} \\$$$${answer}. \\$$
$$\mathrm{maybe}\:\mathrm{u}\:\mathrm{can}\:\mathrm{expand}\:\mathrm{it}\:\mathrm{with}\:\mathrm{Newton}'\mathrm{s}\:\mathrm{binomial} \\$$$$\mathrm{and}\:\mathrm{integrated}\:\mathrm{it}\:\mathrm{manually} \\$$
$$\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{9}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}{C}_{{k}} ^{\mathrm{9}} \mathrm{9}^{{k}} \left({x}^{\mathrm{2}} \right)^{\mathrm{9}−{k}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}{C}_{{k}} ^{\mathrm{9}} \mathrm{9}^{{k}} {x}^{\mathrm{18}−\mathrm{2}{k}} \\$$$$\int\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{9}} {dx}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}{C}_{{k}} ^{\mathrm{9}} \mathrm{9}^{{k}} \int{x}^{\mathrm{18}−\mathrm{2}{k}} {dx} \\$$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\:\frac{\mathrm{9}^{{k}} {C}_{{k}} ^{\mathrm{9}} }{\mathrm{19}−\mathrm{2}{k}}×{x}^{\mathrm{19}−\mathrm{2}{k}} +{C} \\$$