indefinite integral of (1/(e^2x + e^-2x))?

Question

No clue how to solve this!

Answer 1

∫[1/(e^2x + e^-2x)]dx

Now, let
u = e^2x

therefore,
du = 2e^2x dx

but since u = e^2x, then
du = 2u dx

Then
dx = du/2u

We can now substitute
= ∫[1/(u + u‾¹)](du/2u)

We rearrange (distribute the u to the denominator)
= ∫[1/2 du/(u² + 1)]

Then, remove the constant:
= 1/2 ∫ du/(u² + 1)

But ∫ du/(u² + 1) = tan‾¹ u + C, then
= 1/2 tan‾¹ u + C

Now, bring back the x.
Therefore, the integral is
= 1/2 tan‾¹ e^2x + C

^_^

Answer 2

First, let's try eliminating all negative exponents. We'll multiply the top and bottom by e^(2x).

Integral ( [e^(2x)] / (e^2x)[e^(2x) + e^(-2x)] )dx

As a result, e^(2x) times e^(-2x) = 1. Also, e^(2x) times e^(2x) = e^(4x).

Integral ([e^(2x)] / [e^(4x) + 1] )dx

Let's rewrite e^(4x) as [e^(2x)]^2

Integral ([e^(2x)] / [ (e^(2x))^2 + 1] )dx

I'm going to rewrite this slightly to prove a point.

Integral ( [1 / [ (e^(2x))^2 + 1] ] * e^(2x) dx)

To solve this, we need to use substitution.

Let u = e^(2x). Then
du = 2 * e^(2x) dx, and
(1/2) du = e^(2x) dx

If you look at the tail end of the Integral, e^(2x) dx, it's EXACTLY equal to the last step of our substitution, (1/2) du. That is part of our substitution.

Integral [ (1 / [u^2 + 1]) * (1/2) du ]

Let's pull out all constants.

(1/2) * Integral ( 1 / (u^2 + 1) ) du

But that's the derivative of arctan, and one of our known derivatives, so we can solve.

(1/2) * arctan(u^2 + 1) + C

Replacing u = e^(2x), we get

(1/2) * arctan ( [e^(2x)]^2 + 1) + C

Which we can reduce to

(1/2) * arctan (e^(4x) + 1) + C

Answer 3

Simplify it first.

1 / (e^2x + e^-2x) = 1 / (e^2x + 1/(e^2x))

Multiply top and bottom by e^2x, you will obtain

1 / (e^2x + e^-2x) = e^2x / (e^4x + 1)

Notice that it resembles the derivative of arctan(x)
i.e. d[arctan(x)]/dx = 1 / (x^2 + 1)

By substitution,
u = e^2x
du = 2e^2x dx
e^2x dx = du / 2

Therefore,

S( e^2x / (e^4x + 1) ) dx
= S ( (du/2) / (u^2 + 1) )du
= 1/2 S ( du / (u^2 +1) )
= 1/2 arctan(u) + C
= 1/2 arctan(e^2x) + C

The S sign above is the integral sign.

Hope this help!=]

Answer 4

assume e^2x = u then du/dx = e^2x*2
since e^2x = u
du/dx = u*2
dx = du/2*u.
Integral of 1/(u+(1/u))*du/2.u
= du*1/2*(1/u^2+1). At this point you must make use of the inverse trigonometric identities for the derivative of d/dx*tan^-1(x)= 1/(1+x^2),
thus now we have 1/2(tan^-1(u) + C
substitute u with e^2x then you'll get:
1/2*tan^-1(e^2x) + C.

Answer 5

That sucks. Unfortunately, I don't have a clue on how to solve that either. If that's a school math problem. And if someone else can't solve it, you'll just have to skip it. :-(

Answer 6

sin(3a) is carefully a persevering with 'reason there isn't any longer any x in it, so the crucial will be rewritten as ? sin(3a)dx- ?sin(3x)dx ? sin(3a)dx = sin(3a)x for ?sin(3x)dx enable u=3x then du=3dx and dx=a million/3du so ?sin(3x)dx = ?(a million/3)sin(u)du = a million/3?sin(u)du = a million/3cos(u)+c = a million/3cos(3x) +c so the achieved crucial evaluates to (sin3?)x + a million/3 cos3x + C