Question Number 148408 by cesarL last updated on 27/Jul/21

∫x^5 e^x^2 dx Help please!

Answered by Olaf_Thorendsen last updated on 27/Jul/21

F(x) = ∫x^5 e^x^2 dx F(x) = ∫(1/2)x^4 (2xe^x^2 )dx F(x) = (1/2)x^4 e^x^2 −∫2x^3 e^x^2 dx F(x) = (1/2)x^4 e^x^2 −∫x^2 (2xe^x^2 ) dx F(x) = (1/2)x^4 e^x^2 −x^2 e^x^2 +∫2xe^x^2 dx F(x) = (1/2)x^4 e^x^2 −x^2 e^x^2 +e^x^2 +C F(x) = ((1/2)x^4 −x^2 +1)e^x^2 +C

Commented bycesarL last updated on 27/Jul/21

thank u sir

Answered by Math_Freak last updated on 27/Jul/21

∫(x^2 )^2 xe^x^2 dx let u = x^2 dx = (du/(2x)) ∫u^2 xe^u •(du/(2x)) (1/2)∫u^2 e^u du integrating by parts (1/2)(u^2 e^u − 2∫ue^u du) (1/2)(u^2 e^u − 2(ue^u − ∫e^u du)) (1/2)(u^2 e^u − 2(ue^u − e^u )) + C ((u^2 e^u )/2) − ue^u + e^u + C u = x^2 ((x^4 e^x^2 )/2) − x^2 e^x^2 + e^x^2 + C