How can the following integral be solved $\int t^3 e^{t^2+t} dt$?

Question

$$ \int t^3 e^{t^2+t} dt $$Solve the given integral by using the substitution techniques or the integretion by parts

Answer 1

First, we write the integral of interest as

$$\int t^3e^{t^2+1}\,dt=e^{-1/4}\int t^3e^{(t+1/2)^2}\,dt$$

Next, we make the substitution $u=t+1/2$ to obtain

$$\begin{align} \int t^3e^{t^2+1}\,dt&=e^{-1/4}\int (u-1/2)^3e^{u^2}\,du\\\\ &=e^{-1/4}\left(\int u^3e^{u^2}\,du-\frac32\int u^2e^{u^2}\,du+\frac34\int ue^{u^2}\,du-\frac18\int e^{u^2}\,du\right)\tag 1 \end{align}$$

We will use the designation $I_n=\int u^ne^{u^2}\,du$. Then, using integration by parts on $I_3$ and $I_2$, we find

$$\begin{align} I_3&=\frac12(u^2-1)e^{u^2}\\\\ I_2&=\frac12ue^{u^2}-\frac12\int e^{u^2}\,du\\\\ I_1&=\frac12 e^{u^2}\\\\ I_0&=\int e^{u^2}\,du \end{align}$$

To finish, note that the Imaginary Error Function, $\text{erfi}(u)=\frac{2}{\sqrt{\pi}}\int_0^u e^{t^2}\,dt$ so that we can write the indefinite integral

$$\int e^{u^2}\,du=\frac{\sqrt{\pi}}{2}\text{erfi}(u)$$

Finally, we reverse the substitution and replace $u$ with $t+1/2$. The details are left for the reader.