What is #F(x) = int xe^(sqrtx) +x^2 dx# if #F(0) = 1 #?

#F(x)=int(xe^sqrtx+x^2)dx#

First of all, the integral can be split up and the second part integrated rather painlessly:

#F(x)=intxe^sqrtxdx+intx^2dx#

#F(x)=intxe^sqrtxdx+x^3/3#

For the remaining integral, we should first attempt the substitution #t=sqrtx#. Furthermore, we can modify this to also say that #t^2=x#, which when differentiated yields that #2tdt=dx#. Plugging these values into the integral gives us:

#F(x)=intt^2e^t(2tdt)+x^3/3#

#F(x)=2intt^3e^tdt+x^3/3#

To integrate this, we should try to apply integration by parts. Integration by parts takes the form #intudv=uv-intvdu#. In order to make this integral simpler, let the non-exponential piece of this integral, that is, #t^3# be the #u# value so as we differentiate it, it approaches #0# and does not become more complex. Let:

#{(u=t^3" "=>" "du=3t^2dt),(dv=e^tdt" "=>" "v=e^t):}#

Applying these to the integration by parts formula, and distributing the #2#, this becomes:

#F(x)=2[t^3e^t-int3t^2e^tdt]+x^3/3#

#F(x)=2t^3e^t-6intt^2e^tdt+x^3/3#

Perform integration by parts again. The same logic as before will dictate our choice of #u# and #dv#:

#{(u=t^2" "=>" "du=2tdt),(dv=e^tdt" "=>" "v=e^t):}#

So we obtain:

#F(x)=2t^3e^t-6[t^2e^t-int2te^tdt]+x^3/3#

#F(x)=2t^3e^t-6t^2e^t+12intte^tdt+x^3/3#

Once again performing integration by parts:

#{(u=t" "=>" "du=dt),(dv=e^tdt" "=>" "v=e^t):}#

Yielding:

#F(x)=2t^3e^t-6t^2e^t+12[te^t-inte^tdt]+x^3/3#

#F(x)=2t^3e^t-6t^2e^t+12te^t-12inte^tdt+x^3/3#

This integral has already been performed three times previously within the integration by parts:

#F(x)=2t^3e^t-6t^2e^t+12te^t-12e^t+x^3/3+C#

The constant of integration has been added!

Back-substituting with #t=sqrtx#:

#F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+C#

With our initial condition of #F(0)=1# we can determine the exact value of our constant #C#:

#1=2(0)e^0-6(0)e^0+12(0)e^0-12e^0+0/3+C#

Here, note that any term with an #x# or power of #x# will effectively be turned into #0#, and every #e^sqrt0=e^0=1#. Thus, only the #-12e^0=-12# term will remain in this mess.

#1=-12+C#

#C=13#

Thus:

#F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+13#