First, we use substitution.
Let t = sqrt(x) => dt = 1/(2sqrt(x))dxt=√x⇒dt=12√xdx and t^3 = x^(3/2)t3=x32
Then, substituting, we have
intxe^sqrt(x)dx = 2intx^(3/2)e^sqrt(x)1/(2sqrt(x))dx = 2intt^3e^tdt∫xe√xdx=2∫x32e√x12√xdx=2∫t3etdt
Next, we apply integration by parts three times, using the formula
intudv = uv - intvdu∫udv=uv−∫vdu
Integration by Parts 1:
Let u = t^3u=t3 and dv = e^tdtdv=etdt
Then du = 3t^2dtdu=3t2dt and v = e^tv=et
Applying the formula:
2intt^3e^tdt = 2(t^3e^t - 3intt^2e^tdt)2∫t3etdt=2(t3et−3∫t2etdt)
Integration by Parts 2:
Focusing on the remaining integral, let u = t^2u=t2 and dv = e^tdtdv=etdt
Then du = 2tdtdu=2tdt and v = e^tv=et
Applying the formula:
intt^2e^tdt = t^2e^t - 2intte^tdt∫t2etdt=t2et−2∫tetdt
Integration by Parts 3:
Again, focusing on the remaining integral, let u = tu=t and dv=e^tdtdv=etdt
Then du = dtdu=dt and v = e^tv=et
Applying the formula:
intte^tdt = te^t-inte^tdt = te^t - e^t + C∫tetdt=tet−∫etdt=tet−et+C
Substituting our result into the second integration by parts step, we obtain
intt^2e^tdt = t^2e^t-2(te^t-e^t) + C = e^t(t^2-2t+2)+C∫t2etdt=t2et−2(tet−et)+C=et(t2−2t+2)+C
Substituting this into the first integration by parts step, we obtain
2intt^3e^tdt = 2[t^3e^t-3e^t(t^2-2t+2)]+C2∫t3etdt=2[t3et−3et(t2−2t+2)]+C
=2e^t(t^3-3t^2+6t-6)+C=2et(t3−3t2+6t−6)+C
Finally, we substitute t=sqrt(x)t=√x back in to obtain our final result:
intxe^sqrt(x)dx = 2e^sqrt(x)(x^(3/2)-3x+6sqrt(x)-6)+C∫xe√xdx=2e√x(x32−3x+6√x−6)+C
