Integrate \((x^2+9) e^{2x} dx\)

By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
\(\int (x^2+9) \,e^x \,dx=(x^2+9) \int e^{2x} \,dx-\int (x^2+9)’\int e^{2x} \,dx\)
=\(\frac{(x^2+9) \,e^{2x}}{2}-\int 2x \frac{e^{2x}}{2} \,dx\)
Again, applying integration by parts for integrating \(\int \,x \,e^{2x} \,dx\)
\(\int \,x \,e^{2x} \,dx=x\int e^{2x} \,dx-\int (x)’ \int \,e^{2x} \,dx\)
=\(\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx\)
=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\).
?\(\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)
=\(\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})\)
=\(\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C\).