What is the integral of sqrt(4x2+1)?

Question

In words, as you would say it when reading:
What is the integral of the square root of x squared plus 1?
That's so hard to put in characters...I'll give it a try:
........_
(....../..(.....2........)
.)..V...(.4x....+.1.)..dx

The full stops are used because spaces don't appear multiple times in HTML...

This IS NOT homework...

Answer 1

Integral ( sqrt(4x^2 + 1) dx )

To solve this, you need to use trig substitution.

Let x = (1/2)tan(t)
dx = (1/2)sec^2(t) dt.

Skipping details, after the substitution you should obtain

Integral ( sqrt( tan^2(t) + 1 ) (1/2)sec^2(t) dt )

Which should simplify to

(1/2) Integral ( sqrt( sec^2(t) ) sec^2(t) dt )

(1/2) Integral ( sec(t) sec^2(t) dt )

(1/2) Integral ( sec^3(t) dt )

To solve this tricky integral, you have to use parts. Use the information below to get your answer.

∫sec³(x) dx

First, split sec³(x) into sec(x) and sec^2(x).

∫ sec(x) sec²(x) dx

Use integration by parts.

Let u = sec(x). dv = sec²(x) dx
du = sec(x)tan(x) dx. v = tan(x)

sec(x)tan(x) - ∫sec(x)tan²(x)dx

Use the identity tan²(x) = sec²(x) - 1

sec(x)tan(x) - ∫sec(x)[sec²(x) - 1] dx

Distribute the sec(x).

sec(x)tan(x) - ∫(sec³(x) - sec(x)) dx

Separate into two integrals,

sec(x)tan(x) - [∫sec³(x)dx - ∫sec(x)dx]

One of the integrals worth remembering is ∫sec(x)dx = ln|sec(x) + tan(x)|. Substituting that, we get

sec(x)tan(x) - [∫sec³(x)dx - ln|sec(x) + tan(x)|]

Distribute the minus sign.

sec(x)tan(x) - ∫sec³(x) dx + ln|sec(x) + tan(x)|

At this point, it would appear that we've gone in a circle; that isn't the case, because if we look at the whole equation again,

∫sec³(x)dx = sec(x)tan(x) - ∫sec³(x)dx + ln|sec(x) + tan(x)|

What we're going to do at this point is *ADD* ∫sec³(x)dx to both sides as if it were a variable. This eliminates the -∫sec³(x)dx on the right hand side, and makes 2 of them on the left hand side.

2∫sec³(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)|

Now, we multiply both sides by (1/2), to get rid of the 2 on the left hand side.

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|

And, don't forget to add a constant. Our final answer is then:

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C

Answer 2

We know that we have to make 4x2+1 into a single entity, and that since (tany)^2 + 1 = (secy)^2,

Let 4x2 + 1 = (tany)^2 + 1,
=> 4x2 = (tany)^2
=> 2x = tany
=> x = (1/2)tany
=> dx/dy = (1/2)(secy)^2
=> dx = (1/2)(secy)^2 dy

Integral of sqrt(4x2+1) dx
= Integral of sqrt((tany)^2 + 1).(1/2)(secy)^2 dy
= Integral of sqrt(secy)^2.(1/2)(secy)^2 dy
= Integral of (1/2)(secy)^3 dy
:
: I am stuck here, but will keep trying and edit this if I can...

Answer 3

Please do not refer to it as an integral if you're not talking about calculus, but even if you are, integral is the wrong word to use, because it's a term with a specific mean in calculus. But I still don't know what you're trying to ask?

To lav750, you did not give the integral, you gave the solution to finding the derivative.

Answer 4

I = ∫ (4x² + 1) ^(1/2) dx
I = ∫ [ 1² + (2x)² ] ^(1/2) dx
Let u = 2x
du = 2 dx
du / 2 = dx
I = (1/2).∫ [1² + u² ] ^(1/2) du
This is listed as a standard integral given by:-
I = (1/2) sinh^(-1) (u) + C
I = (1/2) sinh^(-1) (2x) + C

Answer 5

Is this ∫√[2/(4x+1)] dx

or ∫√[2/(4x²+1)] dx

or ∫√(4x²+1)dx

or ∫√(x²+1) dx ?

Your question seems to contain all of these

Answer 6

Integral ( sqrt(4x^2 + 1) dx )

To solve this, you need to use trig substitution.

Let x = (1/2)tan(t)
dx = (1/2)sec^2(t) dt.

Skipping details, after the substitution you should obtain

Integral ( sqrt( tan^2(t) + 1 ) (1/2)sec^2(t) dt )

Which should simplify to

(1/2) Integral ( sqrt( sec^2(t) ) sec^2(t) dt )

(1/2) Integral ( sec(t) sec^2(t) dt )

(1/2) Integral ( sec^3(t) dt )

To solve this tricky integral, you have to use parts. Use the information below to get your answer.

∫sec³(x) dx

First, split sec³(x) into sec(x) and sec^2(x).

∫ sec(x) sec²(x) dx

Use integration by parts.

Let u = sec(x). dv = sec²(x) dx
du = sec(x)tan(x) dx. v = tan(x)

sec(x)tan(x) - ∫sec(x)tan²(x)dx

Use the identity tan²(x) = sec²(x) - 1

sec(x)tan(x) - ∫sec(x)[sec²(x) - 1] dx

Distribute the sec(x).

sec(x)tan(x) - ∫(sec³(x) - sec(x)) dx

Separate into two integrals,

sec(x)tan(x) - [∫sec³(x)dx - ∫sec(x)dx]

One of the integrals worth remembering is ∫sec(x)dx = ln|sec(x) + tan(x)|. Substituting that, we get

sec(x)tan(x) - [∫sec³(x)dx - ln|sec(x) + tan(x)|]

Distribute the minus sign.

sec(x)tan(x) - ∫sec³(x) dx + ln|sec(x) + tan(x)|

At this point, it would appear that we've gone in a circle; that isn't the case, because if we look at the whole equation again,

∫sec³(x)dx = sec(x)tan(x) - ∫sec³(x)dx + ln|sec(x) + tan(x)|

What we're going to do at this point is *ADD* ∫sec³(x)dx to both sides as if it were a variable. This eliminates the -∫sec³(x)dx on the right hand side, and makes 2 of them on the left hand side.

2∫sec³(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)|

Now, we multiply both sides by (1/2), to get rid of the 2 on the left hand side.

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|

And, don't forget to add a constant. Our final answer is then:

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C

Answer 7

=sqrt(4x^2+x)=sqrt(5x^2)=5x


(but i have not done calculus for 20 years..lol)

Answer 8

x

Answer 9

9x

Answer 10

we would never do your homeworks for you lol