Question

Integral
please help me

Answer 1

Recall that the product rule for derivatives tells us that, for two functions
`f=f(x)` and
`g=g(x)`,


`(\mathrm d)/(\mathrm dx)[fg]=(\mathrm df)/(\mathrm dx)g+f(\mathrm dg)/(\mathrm dx)`

Integrating both sides with respect to
`x` gives the reverse "rule" for integration:


`\displaystyle\int(\mathrm d)/(\mathrm dx)[fg]\,\mathrm dx=\int(\mathrm df)/(\mathrm dx)g\,\mathrm dx+\int f(\mathrm dg)/(\mathrm dx)\,\mathrm dx`

`\implies \displaystyle\int f(\mathrm dg)/(\mathrm dx)\,\mathrm dx=fg-\int(\mathrm df)/(\mathrm dx)g\,\mathrm dx`

You might know this process by the name "integration by parts". This is the standard method for the given integral.

Take


`f=x\implies(\mathrm df)/(\mathrm dx)=1`

`(\mathrm dg)/(\mathrm dx)=e^(-2x)\implies g=-\frac12e^(-2x)`

and so


`\displaystyle\int xe^(-2x)\,\mathrm dx=-\frac x2e^(-2x)+\frac12\int e^(-2x)\,\mathrm dx`

The remaining integral is trivial, and the overall result is


`\displaystyle\int xe^(-2x)\,\mathrm dx=-\frac x2e^(-2x)-\frac14e^(-2x)+C`