Integrate e^(4x)\sqrt{1+e^(2x) } dx please show work so i can learn

Question

asked Jun 28, 2022 33.1k views

1 Answer

Mrzool · Answer 1 · 2022-07-04T02:18:41+0000

Answer:

$(\frac{(1+e^(2x)) ^{(5)/(2) } }{{5}} + \frac{(1+e^(2x) )^{(3)/(2) } }{{3}} )+C$

Explanation:

Step(i):-

Given that the function

f(x) =
$e^(4x) \sqrt{1+e^(2x) }$

Now integrating on both sides, we get

$\int\limits{f(x)} \, dx = \int\limits{e^(4x) \sqrt{1+e^(2x) } dx$

=
$\int\limits{e^(2x) e^(2x) \sqrt{1+e^(2x) } dx$

Step(ii):-

Let
1 + e^(2x) = t

e^(2x) = t -1

2e^(2x)dx = d t

e^(2x)dx = (1)/(2) d t

=
$\int\limits{( \sqrt{1+e^(2x) }) e^(2x) e^(2x) dx$

=
$\int\limits {√(t)(t-1)(1)/(2) dt }$

=
$(1)/(2) \int\limits {√(t) (t) -√(t) ) dt }$

=
$(1)/(2) \int\limits {(t^{(1)/(2) } t^(1) +t^{(1)/(2) } ) } \, dx$

$= (1)/(2) \int\limits {(t^{(3)/(2) } +t^{(1)/(2) } ) } \, dx$

=
$(1)/(2) (\frac{t^{(3)/(2) +1} }{(3)/(2)+1 } + \frac{t^{(1)/(2) +1} }{(1)/(2)+1 } )+C$

=
$(1)/(2) (\frac{t^{(3)/(2) +1} }{(5)/(2) } + \frac{t^{(1)/(2) +1} }{(3)/(2) } )+C$

=
$(1)/(2) (\frac{t^{(5)/(2) } }{(5)/(2) } + \frac{t^{(3)/(2) } }{(3)/(2) } )+C$

=
$(\frac{(1+e^(2x)) ^{(5)/(2) } }{{5}} + \frac{(1+e^(2x) )^{(3)/(2) } }{{3}} )+C$

Final answer:-

=
$(\frac{(1+e^(2x)) ^{(5)/(2) } }{{5}} + \frac{(1+e^(2x) )^{(3)/(2) } }{{3}} )+C$