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2-e-1-2-x-2-2-dx-

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Question Number 78766 by M±th+et£s last updated on 20/Jan/20
∫2 e^(1/(2(x−2)^2 )) dx
∫2e12(x−2)2dx
Answered by MJS last updated on 20/Jan/20
2∫e^(1/(2(x−2)^2 )) dx= [t=2−x → dx=−dt] =−2∫e^(1/(2t^2 )) dt= by parts: u′=1 → u=t v=e^(1/(2t^2 )) → v′=−(e^(1/(2t^2 )) /t^3 ) =−2te^(1/(2t^2 )) +2∫−(e^(1/(2t^2 )) /t^2 )dt= [u=(1/( (√2)t)) → dt=−(√2)t^2 du] =−2te^(1/(2t^2 )) +(√(2π))∫((2e^u^2 )/( (√π)))du= =−2te^(1/(2t^2 )) +(√(2π))erfi u = =−2te^(1/(2t^2 )) +(√(2π))erfi (1/( (√2)t))= =−2(2−x)e^(1/(2(2−x)^2 )) +(√(2π))erfi ((√2)/(2(2−x))) +C
2∫e12(x−2)2dx=[t=2−x→dx=−dt]=−2∫e12t2dt=byparts:u′=1→u=tv=e12t2→v′=−e12t2t3=−2te12t2+2∫−e12t2t2dt=[u=12t→dt=−2t2du]=−2te12t2+2π∫2eu2πdu==−2te12t2+2πerfiu==−2te12t2+2πerfi12t==−2(2−x)e12(2−x)2+2πerfi22(2−x)+C
Commented by mind is power last updated on 20/Jan/20
Nice
Nice
Commented by M±th+et£s last updated on 20/Jan/20
thank you sir
thankyousir

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