Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com