How do you integrate #int x*2^x# by integration by parts method?

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Answer 1 · 2016-08-03T19:59:20

#=2^x/ (ln 2) ( x - 1/ (ln 2) ) + C#

firstly know that #d/dx ( a ^x ) = ln a \ a^x#

So

#int x*2^x \ dx#

#= int x d/dx( 1/ ln 2 2^x) \ dx#

#= x/ ln 2 2^x - int d/dx(x) 1/ ln 2 2^x\ dx#

#= x/ ln 2 2^x - 1/ ln 2 int 2^x\ dx#

#= x/ ln 2 2^x - 1/ ln 2 int d/dx( 1/ ln 2 2^x)\ dx#

#= x/ ln 2 2^x - 2^x 1/( ln 2 )^2+ C#

#=2^x/ (ln 2) ( x - 1/ (ln 2) ) + C#