Answer
$\displaystyle \frac{(x^2+1)(e^{3x+1})}{2}-\frac{2xe^{3x+1}}{9}-\frac{2}{27} e^{3x+1}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\int (x^2+1) e^{3x+1} \ dx$ and $dv=e^{3x+1} dx$
$\displaystyle \int(x^{2}+1)e^{3x+1}dx=\displaystyle \frac{(x^2+1)(e^{3x+1})}{3}-\frac{2}{3}\int xe^{3x+1} \ dx$
$=\displaystyle \frac{(x^2+1)(e^{3x+1})}{2}-\frac{2xe^{3x+1}}{9}-\frac{2}{9}(\dfrac{1}{3} e^{3x+1})+C$
$=\displaystyle \frac{(x^2+1)(e^{3x+1})}{2}-\frac{2xe^{3x+1}}{9}-\frac{2}{27} e^{3x+1}+C$
