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!

1 Answer
VNVDVI
Apr 21, 2018

#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.

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