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3-x-2-5-x-dx-




Question Number 145259 by mathdanisur last updated on 03/Jul/21
∫ (((3(√x)+2)^5 )/( (√x))) dx = ?
$$\int\:\frac{\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{5}} }{\:\sqrt{{x}}}\:{dx}\:=\:? \\ $$
Answered by Dwaipayan Shikari last updated on 03/Jul/21
put (√x)=t⇒(1/(2(√x)))dx=dt  =2.3^5 ∫(t+(2/3))^5 dt  =3^4 (t+(2/3))^6 +C=81((√x)+(2/3))^6 +C
$${put}\:\sqrt{{x}}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}={dt} \\ $$$$=\mathrm{2}.\mathrm{3}^{\mathrm{5}} \int\left({t}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{5}} {dt} \\ $$$$=\mathrm{3}^{\mathrm{4}} \left({t}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{6}} +{C}=\mathrm{81}\left(\sqrt{{x}}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{6}} +{C} \\ $$
Commented by mathdanisur last updated on 03/Jul/21
thanks Ser  Answer: (1/(81))(3(√x)+2)^6  or (1/9)(3(√x)+2)^6  ?
$${thanks}\:{Ser} \\ $$$${Answer}:\:\frac{\mathrm{1}}{\mathrm{81}}\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{6}} \:\boldsymbol{{or}}\:\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{6}} \:? \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jul/21
(1/9)(3(√x)+2)^6 +C
$$\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{6}} +{C} \\ $$
Commented by mathdanisur last updated on 03/Jul/21
Ser, but in your solution 81((√x)+(2/3))^6
$${Ser},\:{but}\:{in}\:{your}\:{solution}\:\mathrm{81}\left(\sqrt{{x}}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{6}} \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jul/21
81((√x)+(2/3))^6 =(3^4 /3^6 )(3(√x)+2)^6 =(((3(√x)+2)^6 )/9)
$$\mathrm{81}\left(\sqrt{{x}}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{6}} =\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{6}} =\frac{\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{6}} }{\mathrm{9}} \\ $$
Commented by mathdanisur last updated on 03/Jul/21
Thanks Ser
$${Thanks}\:{Ser} \\ $$

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