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Question

Solve : 10eex(1+x.ex)dx

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Solution

10eex(1+xex)dx

substitute u=ex

eu(1+ln(u)u)udu

euu+euln(u)

Ei(u)+euln(u)Ei(u)

euln(u)

eexln(ex)

eexx+C

10eex(1+xex)dx

ee0

ee

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