23. $\\int x^{3}e^{-2x}dx$

Explanation:

Step1: Use integration by parts formula ( \int u dv=uv-\int v du )

Let ( u = x^{3}), (dv=e^{-2x}dx). Then (du = 3x^{2}dx), (v=-\frac{1}{2}e^{-2x}) [ \begin{align*} \int x^{3}e^{-2x}dx&=-\frac{1}{2}x^{3}e^{-2x}+\frac{3}{2}\int x^{2}e^{-2x}dx \end{align*} ]

Step2: Apply integration by parts again on ( \int x^{2}e^{-2x}dx )

Let (u = x^{2}), (dv=e^{-2x}dx). Then (du = 2xdx), (v =-\frac{1}{2}e^{-2x}) [ \begin{align*} \int x^{2}e^{-2x}dx&=-\frac{1}{2}x^{2}e^{-2x}+\int xe^{-2x}dx \end{align*} ]

Step3: Apply integration by parts once more on ( \int xe^{-2x}dx )

Let (u = x), (dv=e^{-2x}dx). Then (du = dx), (v=-\frac{1}{2}e^{-2x}) [ \begin{align*} \int xe^{-2x}dx&=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx \end{align*} ]

Step4: Integrate ( \int e^{-2x}dx )

[ \int e^{-2x}dx=-\frac{1}{2}e^{-2x}+C ]

Step5: Substitute back all the results

[ \begin{align*} \int x^{3}e^{-2x}dx&=-\frac{1}{2}x^{3}e^{-2x}+\frac{3}{2}\left(-\frac{1}{2}x^{2}e^{-2x}+\left(-\frac{1}{2}xe^{-2x}+\frac{1}{2}\left(-\frac{1}{2}e^{-2x}\right)\right)\right)+C\ &=-\frac{1}{2}x^{3}e^{-2x}-\frac{3}{4}x^{2}e^{-2x}-\frac{3}{4}xe^{-2x}-\frac{3}{8}e^{-2x}+C \end{align*} ]

Answer:

(-\frac{1}{2}x^{3}e^{-2x}-\frac{3}{4}x^{2}e^{-2x}-\frac{3}{4}xe^{-2x}-\frac{3}{8}e^{-2x}+C)