What is ∫ e^ x (x ^3 x- x)/( ^2 x) d x equal to?

Let us differentiate all the options one by one to get the expression in the question whose integral is to be found. Here $x e^{\sin x}$ is the common term in all the options. So, let us differentiate it first. $$\begin{aligned} & \text{ Let } l=x e^{\sin x} \\ & \Rightarrow \frac{d l}{d x}=e^{\sin x}[x \cos x+1] \\ & \Rightarrow \frac{d l}{d x}=\frac{e^{\sin x}}{\cos ^2 x}\left[x \cos ^3 x+\cos ^2 x\right] \end{aligned}$$ Let $m=\sec x e^{\sin x}$$\begin{aligned} & \Rightarrow \frac{d m}{d x}=\sec x e^{\sin x} \cdot \cos x+e^{\sin x} \sec x \tan x \\ & \Rightarrow \frac{d m}{d x}=e^{\sin x}\left[1+\frac{\sin x}{\cos ^2 x}\right] \\ & \Rightarrow \frac{d m}{d x}=\frac{e^{\sin x}}{\cos ^2 x}\left[\cos ^2 x+\sin x\right] \end{aligned}$$$Differentiation of option (a) is$$\begin{aligned} & =\frac{e^{\sin x}}{\cos ^2 x}\left[x \cos ^3 x+\cos ^2 x+\cos ^2 x+\sin x\right] \\ & =\frac{e^{\sin x}}{\cos ^2 x}\left[x \cos ^3 x+2 \cos ^2 x+\sin x\right] \end{aligned}$$Differentiation of option (b) is$$\begin{aligned} & =\frac{e^{\sin x}}{\cos ^2 x}\left[x \cos ^3 x+\cos ^2 x-\cos ^2 x-\sin x\right] \\ & =\frac{e^{\sin x}}{\cos ^2 x}\left[x \cos ^3 x-\sin x\right] \end{aligned}$$\therefore$ Option (2) is correct.