How to evaluate 21(e2xex1) ? Can you solve it without substitution and somehow use chain rule?

1 Answer
Jun 14, 2018

21e2xex1dx=e(e1)+ln(e+1)

Explanation:

21e2xex1dx=21exex1exdx

21e2xex1dx=21exex1d(ex)

21e2xex1dx=21ex1+1ex1d(ex)

21e2xex1dx=21(1+1ex1)d(ex)

21e2xex1dx=21d(ex)+211ex1d(ex)

21e2xex1dx=21d(ex)+21d(ex1)ex1

21e2xex1dx=[ex+ln(ex1)]21

21e2xex1dx=e2+ln(e21)eln(e1)

21e2xex1dx=e(e1)+ln(e21e1)

21e2xex1dx=e(e1)+ln(e+1)