Problem in indefinite integral. (Exponential)

Question

I'm given this integral to integrate. I've no idea where to start with. Perhaps someone can give me some hints or guide me. Thanks a lot.

$$\int\frac{(x^3)e^{x^2}{}}{x^2+1}dx$$

Answer 1

Let $u=1+x^2$. Then, $du=2xdx$, and $x^2=u-1$. Thus, we have

$$\begin{align} \int \frac{x^3e^{x^{2}}}{1+x^2}dx&=\int \frac{x^2e^{x^{2}}}{1+x^2}xdx\\\\ &=\frac12 \int \frac{(u-1)e^{u-1}}{u}du\\\\ &=\frac{1}{2e}\left(e^u-\int \frac{e^u}{u}du\right)\\\\ &=\frac{1}{2e}\left(e^u-Ei(u)\right)\\\\ &=\frac{1}{2e}\left(e^{1+x^2}-Ei(1+x^2)\right)+C_1 \end{align}$$

We can expand this result in a Taylor series, say about $u=1$ (i.e., $x=0$). We proceed to write the first four derivatives of $Ei(u)$ as

$$\begin{align} Ei^{1}(u) &= \left(\frac{1}{u}\right)e^u\\ Ei^{2}(u)&=\left(\frac{1}{u}-\frac{1}{u^2}\right)e^u\\ Ei^{3}(u)&=\left(\frac{1}{u}-\frac{2}{u^2}+\frac{2}{u^3}\right)e^u\\ Ei^{4}(u)&=\left(\frac{1}{u}-\frac{3}{u^2}+\frac{6}{u^3} -\frac{6}{u^4}\right)e^u\\ \end{align}$$

whereupon evaluating at $u=1$ gives

$$\begin{align} Ei(u)&=Ei(1)+e\left((u-1)+\frac16 (u-1)^3-\frac{1}{12}(u-1)^4\right)+O(u^5)\\\\ &=Ei(1)+e\left(x^2+\frac16 x^6-\frac{1}{12}x^8\right)+O(x^{10}) \end{align}$$

Recalling that the series for $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$, we have

$$\begin{align} \int \frac{x^3e^{x^{2}}}{1+x^2}dx&=\frac{1}{2e}\left(e^{1+x^2}-Ei(1+x^2)\right)+C_1\\\\ &=\frac{1}{4} \left(x^4-\frac{1}{12} x^8\right)+O(x^{10})+C_2 \end{align}$$

Answer 2

We have $$\int\frac{x^{3}e^{x^{2}}}{x^{2}+1}dx=\int\left(xe^{x^{2}}-\frac{xe^{x^{2}}}{x^{2}+1}\right)dx=\frac{1}{2}e^{x^{2}}-\frac{\textrm{Ei}\left(x^{2}+1\right)}{2e}+C $$ where $\textrm{Ei}\left(x^{2}+1\right) $ is the exponential integral. Maybe it's useful to remaind that $$\frac{d}{dx}\textrm{Ei}\left(x\right)=\frac{e^{x}}{x}. $$