Question

Find the integral, using techniques from this or the previous chapter.
∫(3x+6)e-3x dx.

Answer 1

Integration of the expression will be
`I=e^(-3x)(1-x)-2e^(-3x)+c`

Explanation:

We have given expression
`\int (3x+6)e^(-3x)dx`

`\int (3x+6)e^(-3x)dx=\int (3xe^(-3x)+6e^(-3x))dx`

Let
`I=I_1+I_2`, here
`I_1=3xe^(-3x)` and
`I_2=6e^(-3x)`

`I_1=\int 3xe^(-3x)`

Integrating by part

`I_1=3[x\int e^(-3x)-\int e^(-3x)(d)/(dx)x]`

`I_1=3[x (e^(-3x))/(-3)+(e^(-3x))/(3)]`+c

`I_1=e^(-3x)(1-x)`+c

Now
`I_2=\int 6e^(-3x)dx`

`I_2= 6* (e^(-3x))/(-3)+c`

`I_2=-2e^(-3x)+c`

So integration I will be equal to

`I=e^(-3x)(1-x)-2e^(-3x)+c`