Trying to solve this tricky one: \(\displaystyle\int{e}^{{{2}{x}}}\sqrt{{{1}+{e}^{{{2}{x}}}}}{\left.{d}{x}\right.}\)

Deja Vaughn

Deja Vaughn

Answered question

2022-04-09

Trying to solve this tricky one:
e2x1+e2xdx

Answer & Explanation

Mey9ci0

Mey9ci0

Beginner2022-04-10Added 14 answers

Let u=1+e2x, then du=2e2xdx, that is e2xdx=12du. Using this, the integral can be written as
{e2x}1+e2xdx=u12du
=13u32+C
=13(1+e2x)32+C

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