Remember that #\int\ f(x)+-g(x)\ dx=\int\ f(x)\ dx+-\int\ g(x) dx#.
We can separate the integral into #\int\ xe^(x^2-1)\ dx-\int\ x^2e^x\ dx#.
First Integral: #\int\ xe^(x^2-1)\ dx#
For the first integral, we apply a substitution of #u=x^2-1#. Then, #du=2x\ dx#, or #dx=(du)/(2x)#.
We substitute these values into the integral: #\int\ (xe^u\ du)/(2x)=\int\ e^u/2\ du=e^u/2+C#. Substituting #u=x^2-1# in, we get #e^(x^2-1)/2+C#.
Second Integral: #\int\ x^2e^x\ dx#
We apply integration by parts here: #\int\ u\ dv=uv-\int\ v\ du#. Following the LIATE rule, we set #u=x^2# and #dv=e^x#. Then, #du=2x# and #v=e^x#.
Substituting these in, we get #\int\ x^2e^x\ dx=x^2e^x-\int\ e^x2x\ dx#. We use integration by parts again. This time, #u=2x#, #dv=e^x#, #du=2#, and #v=e^x#.
Then, #\int\ e^x\2x\ dx=2xe^x-\int\ 2e^x\ dx=2xe^x-2e^x+C#.
We substitute this back in: #\int\ x^2e^x\ dx=x^2e^x-\int\ e^x2x\ dx=x^2e^x-2xe^x+2e^x+C#.
Finally, we bring everything back together: #\int\ xe^(x^2-1)\ dx-\int\ x^2e^x\ dx=e^(x^2-1)/2-x^2e^x+2xe^x-2e^x+C=e^(x^2-1)/2-e^x(x^2-2x+2)+C#.
We are given one condition that f(2)=7. Therefore, when #x=2#, #e^(x^2-1)/2-e^x(x^2-2x+2)+C=7#. We need to find the constant #C# such that #-2e^2+e^3/2+C=7#. Solving this, we get that #C=7+2e^2-e^3/2#.
Therefore, #f(x)=e^(x^2-1)/2-e^x(x^2-2x+2)+2e^2-e^3/2+7#.
