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Question Number 29455 by prof Abdo imad last updated on 08/Feb/18
find ∫ 3^(√(2x+1)) dx .
$${find}\:\int\:\:\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}} \:{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 12/Feb/18
the ch. (√(2x+1)) =t give 2x+1=t^2 ⇒x=(1/2)t^2 −(1/2) I= ∫3^t tdt = ∫ t e^(tln(3)) dt by parts I = (1/(ln3))t e^(tln3) −∫ (1/(ln3)) e^(tln3) dt =((t 3^t )/(ln3)) − (1/((ln3)^2 ))e^(tln(3)) +k =(((√(2x+1)) 3^(√(2x+1)) )/(ln3)) −(1/((ln3)^2 ))3^((√(2x+1)) ) +k .
$${the}\:{ch}.\:\sqrt{\mathrm{2}{x}+\mathrm{1}}\:={t}\:{give}\:\mathrm{2}{x}+\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}=\:\int\mathrm{3}^{{t}} \:{tdt}\:=\:\int\:{t}\:{e}^{{tln}\left(\mathrm{3}\right)} {dt}\:{by}\:{parts} \\ $$$${I}\:=\:\frac{\mathrm{1}}{{ln}\mathrm{3}}{t}\:{e}^{{tln}\mathrm{3}} \:\:−\int\:\frac{\mathrm{1}}{{ln}\mathrm{3}}\:{e}^{{tln}\mathrm{3}} {dt} \\ $$$$=\frac{{t}\:\mathrm{3}^{{t}} }{{ln}\mathrm{3}}\:−\:\frac{\mathrm{1}}{\left({ln}\mathrm{3}\right)^{\mathrm{2}} }{e}^{{tln}\left(\mathrm{3}\right)} \:\:+{k} \\ $$$$=\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}\:\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}} }{{ln}\mathrm{3}}\:−\frac{\mathrm{1}}{\left({ln}\mathrm{3}\right)^{\mathrm{2}} }\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}\:} \:\:\:+{k}\:. \\ $$
Answered by sma3l2996 last updated on 09/Feb/18
A=∫3^(√(2x+1)) dx=∫e^((√(2x+1))ln(3)) dx =∫((√(2x+1))/(ln(3)))×((ln(3))/( (√(2x+1))))e^((√(2x+1))ln(3)) dx u=(√(2x+1))⇒u′=(1/( (√(2x+1)))) v′=((ln(3))/( (√(2x+1))))e^((√(2x+1))ln(3)) ⇒v=e^((√(2x+1))ln(3)) A=(1/(ln(3)))(√(2x+1))e^((√(2x+1))ln(3)) −(1/(ln3))∫(e^((√(2x+1))ln3) /( (√(2x+1))))dx+c =((√(2x+1))/(ln3))×3^(√(2x+1)) −(1/((ln3)^2 ))e^((√(2x+1))ln3) +C A=(1/((ln3)^2 ))(ln(3)(√(2x+1))−1)3^(√(2x+1)) +C
$${A}=\int\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}} {dx}=\int{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\left(\mathrm{3}\right)} {dx} \\ $$$$=\int\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{{ln}\left(\mathrm{3}\right)}×\frac{{ln}\left(\mathrm{3}\right)}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\left(\mathrm{3}\right)} {dx} \\ $$$${u}=\sqrt{\mathrm{2}{x}+\mathrm{1}}\Rightarrow{u}'=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}} \\ $$$${v}'=\frac{{ln}\left(\mathrm{3}\right)}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\left(\mathrm{3}\right)} \Rightarrow{v}={e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\left(\mathrm{3}\right)} \\ $$$${A}=\frac{\mathrm{1}}{{ln}\left(\mathrm{3}\right)}\sqrt{\mathrm{2}{x}+\mathrm{1}}{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\left(\mathrm{3}\right)} −\frac{\mathrm{1}}{{ln}\mathrm{3}}\int\frac{{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\mathrm{3}} }{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}{dx}+{c} \\ $$$$=\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{{ln}\mathrm{3}}×\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}} −\frac{\mathrm{1}}{\left({ln}\mathrm{3}\right)^{\mathrm{2}} }{e}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}{ln}\mathrm{3}} +{C} \\ $$$${A}=\frac{\mathrm{1}}{\left({ln}\mathrm{3}\right)^{\mathrm{2}} }\left({ln}\left(\mathrm{3}\right)\sqrt{\mathrm{2}{x}+\mathrm{1}}−\mathrm{1}\right)\mathrm{3}^{\sqrt{\mathrm{2}{x}+\mathrm{1}}} +{C} \\ $$

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