Question

How do you find `intx^2 e^(1-x)dx` using integration by parts?

`intx^2 e^(1-x)dx`

The answer is supposedly `-x^2 e^(1-x) -2xe^(1-x) -2e^(1-x) + C`, but I don't understand how it came to be.

Please explain. Thanks in advance!

Answer 1

`intx^2e^(1-x)dx=-x^2e^(1-x)-2xe^(1-x)-2e^(1-x)+C`

Explanation:

We'll make the following selections:

`u=x^2`

Take its differential:

`du=2xdx`

`dv=e^(1-x)dx`

Take its integral:

`v=inte^(1-x)dx=-e^(1-x)`

Apply the integration by parts formula.

`uv-intvdu=-x^2e^(1-x)-int-2xe^(1-x)dx`

The negatives cancel each other out.

`=-x^2e^(1-x)+int2xe^(1-x)dx`

For `int2xe^(1-x)dx` we're going to have to apply integration by parts again, making the following selections:

`u=2x`

`du=2dx`

`dv=e^(1-x)dx`

`v=-e^(1-x)`

`uv-intvdu=-2xe^(1-x)+2inte^(1-x)dx`

`inte^(1-x)dx=-e^(1-x)`, so we get

`int2xe^(1-x)dx=-2xe^(1-x)-2e^(1-x)`

Thus, the entire integral becomes

`intx^2e^(1-x)dx=-x^2e^(1-x)-2xe^(1-x)-2e^(1-x)+C`, adding in the constant of integration.