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

[Hamza Mahmood]

Evaluate the indefinite integral:


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

[galactus]

I see no one has posted a reply to this one yet. I managed to get it by using 'parts'.

First parts:

Let $$u=\frac{\sin(x)}{x+1}, \;\ du=\frac{(x+1)\cos(x)-\sin(x)}{(x+1)^{2}}dx, \;\ dv=\frac{e^{-x}(x^{2}+3x+3)}{(x+1)^{2}}dx, \;\ v=\frac{-e^{-x}(x+2)}{x+1}$$

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

For the last integral, let $$u=e^{-x}(x+2)(x+1)\cos(x)-e^{-x}(x+2)\sin(x), \;\ du=e^{-x}(x+1)^{2}(\cos(x)+\sin(x))dx, \;\ v=\frac{-1}{2(x+1)^{2}}, \;\ dv=\frac{1}{(x+1)^{3}}dx$$

$$\frac{-1}{2(x+1)^{2}}\left(e^{-x}(x+1)(x+2)\cos(x)-e^{-x}(x+2)\sin(x)\right)-\int\frac{1}{2(x+1)^{2}}\left(e^{-x}(x+1)^{2}(\cos(x)+\sin(x))\right)dx$$

This last integral is easy because the $(x+1)^{2}$ term cancel. Putting this together with the left part of (1) results in:

$$\frac{-e^{-x}(x+2)\sin(x)}{(x+1)^{2}}-\frac{e^{-x}((x+1)\cos(x)-(x+2)\sin(x))}{2(x+1)^{2}}$$

$$=\boxed{\frac{-e^{-x}((x+1)\cos(x)+(x+2)\sin(x))}{2(x+1)^{2}}}$$