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Qn-e-2x-2-2x-6-dx-I-need-help-plz-

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Question Number 73668 by liki last updated on 14/Nov/19
Qn . ∫(e^(2x^2 +2x+6) )dx ... I need help plz..
$$\:{Qn}\:.\:\int\left({e}^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}} \right){dx} \\ $$$$\:…\:{I}\:{need}\:{help}\:{plz}.. \\ $$$$ \\ $$$$ \\ $$
Answered by arkanmath7@gmail.com last updated on 14/Nov/19
= ∫e^(2x^2 +2x ) . e^6 dx = e^6 ∫e^(2x^2 +2x ) dx = e^6 ∫e^(2x^2 +2x+(1/2)−(1/2) ) dx = e^6 ∫e^(((√2)x+(1/( (√2))))^2 −(1/2) ) dx =e^(−(1/2)) e^6 ∫e^((((2x+1)/( (√2))))^2 ) dx now let u = ((2x+1)/( (√2))) ⇒ du = (√(2 ))dx ⇒ du = (√(2 ))dx ⇒ dx = (1/( (√2))) du ∫e^(2x^2 +2x+6) dx = e^((11)/2) . (1/( (√2))) ∫e^u^2 du = (((√π)e^((11)/2) )/(2(√2))) ∫((2e^u^2 )/( (√π)))du error function = (((√π)e^((11)/2) )/(2(√2))) erf(u) = (((√π)e^((11)/2) )/(2(√2))) erf(((2x+1)/( (√2)))) + c
$$=\:\int{e}^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\:} .\:{e}^{\mathrm{6}} \:{dx} \\ $$$$=\:{e}^{\mathrm{6}} \int{e}^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\:} {dx} \\ $$$$=\:{e}^{\mathrm{6}} \int{e}^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:} {dx}\: \\ $$$$=\:{e}^{\mathrm{6}} \int{e}^{\left(\sqrt{\mathrm{2}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\:} {dx}\: \\ $$$$={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{\mathrm{6}} \int{e}^{\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:} {dx}\: \\ $$$$ \\ $$$${now}\:{let}\:{u}\:=\:\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\Rightarrow\:{du}\:=\:\sqrt{\mathrm{2}\:}{dx} \\ $$$$\:\:\Rightarrow\:{du}\:=\:\sqrt{\mathrm{2}\:}{dx}\:\Rightarrow\:{dx}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{du} \\ $$$$\int{e}^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}} {dx}\:=\:{e}^{\frac{\mathrm{11}}{\mathrm{2}}} \:.\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int{e}^{{u}^{\mathrm{2}} } {du} \\ $$$$\:=\:\:\frac{\sqrt{\pi}{e}^{\frac{\mathrm{11}}{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\frac{\mathrm{2}{e}^{{u}^{\mathrm{2}} } }{\:\sqrt{\pi}}{du}\:\:\:\:\:{error}\:{function} \\ $$$$\:=\:\:\frac{\sqrt{\pi}{e}^{\frac{\mathrm{11}}{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:{erf}\left({u}\right) \\ $$$$\:=\:\:\frac{\sqrt{\pi}{e}^{\frac{\mathrm{11}}{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:{erf}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:+\:{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by liki last updated on 14/Nov/19
Thanks so much..
$$\:\:{Thanks}\:{so}\:{much}.. \\ $$
Commented by liki last updated on 14/Nov/19
well soln

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