How do you integrate by parts #(x*e^(4x)) dx#?

1 Answer
Stefan V.
Aug 29, 2015

#int(x * e^(4x)dx) = 1/16 * e^(4x) * (4x-1) + c#

Explanation:

Remember that the formula that allows you to integrate a function by parts looks like this

#color(blue)(int(u * dv) = u * v - int(v * du))" "#, where

#u#, #v# - are two functions of #x#;
#du#, #dv# - their derivatives.

So, you need to identify #u# and #dv#, then calculate #du# and #v#.

If you take #u = x# and #dv = e^(4x)#, you will have

#u = x implies du = dx#

and

#dv = e^(4x) implies v = int(e^(4x)dx) = 1/4 * e^(4x)#

Your target integral will thus be

#int(x * e^(4x)dx) = x * 1/4 * e^(4x) - int(1/4 e^(4x) * dx)#

#int(x * e^(4x)dx) = 1/4 * x * e^(4x) - 1/4 * 1/4 * e^(4x) + c#

#int(x * e^(4x)dx) = 1/4e^(4x)(x - 1/4) + c#

#int(x * e^(4x)dx) = color(green)(1/16 * e^(4x) * (4x-1) + c)#

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