Question #7280b

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Answer 1 · 2018-01-02T08:37:52

#2xe^x-2e^x+C#

or

#2e^x(x-1) + C#

To find #int 2x e^xdx# use integration by parts.

The formula for Integration by Parts is: #intudv =uv-intvdu#

First just factor out the 2 because it's a constant multiple:

#2intxe^xdx#.

Let #u=x\rightarrow du = dx#
Let #dv = e^xdx\rightarrow v=e^x#

#2intxe^xdx = 2(xe^x-inte^xdx)#

#=2xe^x-2e^x+C#

You might factor that a bit:

#=2xe^x-2e^x+C = 2e^x(x-1) + C#