# Use the reduction formulas in a table of integrals to evaluate the following integrals. int x^3e^{2x}dx

Question
Integrals
Use the reduction formulas in a table of integrals to evaluate the following integrals.
$$\int x^3e^{2x}dx$$

2020-11-10
$$\text{Given }I=\int x^3e^{2x}dx$$
$$\text{For evaluating given integral, we use integral by parts theorem}$$
$$\text{According to integral by parts theorem}$$
$$\int f(x)g'(x)dx=f(x)\int g'(x)dx-[\int f'(x)\int g'(x)dx)dx]$$
$$\text{Here, }f(x)=x^3,g'(x)=e^{2x}$$
$$\text{So, be using equation}$$
$$I=\int x^3e^{2x}dx$$
$$=x^3\int e^{2x}dx-[\int(\frac{d}{dx}(x^3)\int e^{2x}dx)dx]$$
$$(\because \int e^{kx}dx=\frac{e^{kx}}{k}+c,\frac{d}{dx}(x^n)=(nx^{n-1})$$
$$=x^3(\frac{e^{2x}}{2})-[\int (3x^2)(\frac{e^{2x}}{2}dx]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3}{2}[\int x^2e^{2x}dx]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3}{2}[x^2\int e^{2x}dx-[\int[\frac{d}{dx}(x^2)\int e^{2x}dx]dx]]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3}{2}[x^2(\frac{e^{2x}}{2})-[\int(2x)\frac{e^{2x}}{2}dx]]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3x^2e^{2x}}{4}+\frac{3}{2}\int xe^{2x}dx$$
$$=\frac{x^3e^{2x}}{2}-\frac{3x^2e^{2x}}{4}+\frac{3}{2}[x\int e^{2x}dx-[\int(\frac{d}{dx}(x)\int e^{2x}dx)dx]]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3x^2e^{2x}}{4}+\frac{3}{2}[x(\frac{e^{2x}}{2}-[\int(1)(\frac{e^{2x}}{2})dx]]$$
$$=\frac{x^3e^{2x}}{2}-\frac{3x^2e^{2x}}{4}+\frac{3xe^{2x}}{4}-\frac{3e^{2x}}{8}+c$$
$$=\frac{(4x^3-6x^2+6x-3)e^{2x}}{8}+c$$
$$\text{Hence, given integral is equal to } \frac{(4x^3-6x^2+6x-3)e^{2x}}{8}+c$$

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