Question

Evaluate the integral using integration by parts where possible.
(x^2 + 1)e^(−2x) dx

Answer 1

\[\begin{array}{l} \text{Expanding the integrand }e^{-2 x} \left(x^2+1\right)\text{ gives }e^{-2 x} x^2+e^{-2 x}: \\ \text{}=\int\limits \left(e^{-2 x} x^2+e^{-2 x}\right) \, dx \\ \text{Integrate the \sum term by term:} \\ \text{}=\int\limits e^{-2 x} x^2 \, dx+\int\limits e^{-2 x} \, dx \\ \text{For the integrand }e^{-2 x} x^2\text{, integrate by parts, }\int\limits f \, dg=f g-\int\limits g \, df\text{, where $\backslash $n }f=x^2\text{, }d@! g=e^{-2 x}d@! x\text{,$\backslash $n}d@! f=2 xd@! x\text{, }g=-\frac{1}{2} e^{-2 x}: \\ \text{}=-\frac{1}{2} e^{-2 x} x^2+\int\limits e^{-2 x} \, dx+\int\limits e^{-2 x} x \, dx \\ \text{For the integrand }e^{-2 x} x\text{, integrate by parts, }\int\limits f \, dg=f g-\int\limits g \, df\text{, where $\backslash $n }f=x\text{, }d@! g=e^{-2 x}d@! x\text{,$\backslash $n}d@! f=\text{}d@! x\text{, }g=-\frac{1}{2} e^{-2 x}: \\ \text{}=-\frac{1}{2} e^{-2 x} x^2-\frac{1}{2} e^{-2 x} x+\frac{3}{2}\int\limits e^{-2 x} \, dx \\ \text{The \int\limits of }e^{-2 x}\text{ is }-\frac{1}{2} e^{-2 x}: \\ \text{}=-\frac{1}{2} e^{-2 x} x^2-\frac{1}{2} e^{-2 x} x-\frac{e^{-2 x}}{4}+\int\limits e^{-2 x} \, dx \\ \text{The \int\limits of }e^{-2 x}\text{ is }-\frac{1}{2} e^{-2 x}: \\ \text{}=-\frac{1}{2} e^{-2 x} x^2-\frac{1}{2} e^{-2 x} x-\frac{3 e^{-2 x}}{4}+\text{constant} \\ \text{Which is equal \to:} \\ \text{}=-\frac{1}{4} e^{-2 x} \left(2 x^2+2 x+3\right)+\text{constant} \\\end{array}\]