Help with taking the integral of a calc problem!!!?

Question

Ok so I have to take the integral of this problem

(5) / square root of (2+4x)

(the square root is all of 2+4x)


any help would be appreciated! thanks

Answer 1

∫ 5/sqrt(2 + 4x) dx
You can use a u-substitution if it will help, but it's not necessary
u = 2 + 4x
du = 4 dx or du/4 = dx
∫ 5/sqrt(2 + 4x) dx = ∫ 5/sqrt(u) (du/4)
= ∫ (5/4)u^(-1/2) du
= (5/4) ∫ u^(-1/2) du
= (5/4)(2u^(1/2)) + C
= (5/2)u^(1/2)

now resubstitute:
(5/2)u^(1/2) = (5/2)(2 + 4x)^(1/2)
answer:
(5/2)sqrt(2 + 4x) + C

Answer 2

Here you go...

∫[ 5 / √(2+4x) ] dx

First of all, we can pull the constant term of 5 to the outside:

5 ∫[ 1 / √(2+4x) ] dx

Now, let's go with u-substitution. Let:

u = 2 + 4x
du = 4dx ==> dx = du/4

Therefore, if we plug this in, we get:

5 ∫[ 1 /√(u) ] du/4

Now, we can pull out a constant of 1/4 and rewrite the square root in the denominator as:

(5/4) ∫[ u^(-1/2) ] dx

So, we just raise this exponent by one and divide by that exponent, so the next exponent is 1/2, and dividing by 1/2 is the same as multiplying by 2, so...

(5/4)*[ 2u^(1/2) ] + C

Pull out the factor of 2, while cancels a 2 in the denominator:

(5/2)*[ u^(1/2) ] + C

Now, just plug in 2 + 4x for u and rewrite the 1/2 exponent as a radical, giving you the final answer:

(5/2)√(2 + 4x) + C
______________

Answer 3

okay so integral is : int 5/(2+4x)^1/2 is as follows
use u-substitution. so u=2+4x
du/dx=4 (derive what you substituted for u)
1/4du=dx (get all your constants on the side of du)
Now you're ready to integrate, your integral looks like the following: 5*1/4 int (du/u^1/2) (NOTE that you're supposed to take all constants outside your integral)

Next: 5/4 (2u^1/2)du (now you're ready to substitute back into the solved integral, ie: your u)
so, 5/4 2(2+4x)^1/2
Simplify to get 5/2 (2+4x)^1/2 +C (your final answer)

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This site show you how to do derivatives step by step and also how to do integrals step by step.
But you have to pay for membership if you want to see the step by step results for integrals but derivatives are for free.

Answer 4

Integral of 5/sqrt(2+4x) dx
= Integral of 5(2+4x)^(-1/2) dx
= [5(2+4x)^(1/2)] / (1/2)(4)
= [5sqrt(2+4x)] / 2
= (5/2)sqrt(2+4x)

Answer 5

you utilize substitution for those.. you regularly have one bit it is complicated, yet could be substituted with something extra convenient that is integrated easily. Your by-product (dx) additionally modifications. right here, have a glance: First one: (d is by-product and int potential fundamental, ok?) Now, as you will locate, it rather is the denominator it is making issues complicated. So enable's attempt substituting it with something extra convenient. enable u=5+e^x Taking the by-product of this substituted equation => du=e^x dx that's an element of the numerator of your query so which you would be able to write the query like this: => int [4e^x /( 5+e^x) dx] = int [4/u du] = 4 ln (u) + consistent = 4 ln (5+e^x) + consistent that's your very final answer. do no longer forget approximately to resubstitute your assumption! _______________________________ 2d one: enable u = (x^2 + 6x + 8) => du= (2x+6) dx => du/2 = (x+3) dx that's an element of the numerator of your query, back. so which you would be able to write that as... => Int[du/(2u)] =0.5 ln (u)+ const. =0.5 ln (x^2 + 6x + 8) + const. it is all! wish it facilitates.

Answer 6

I = ∫ 5 / (2 + 4x)^(1/2).dx
I = ∫ 5.(2 + 4x)^(-1/2).dx
Let u = 2 + 4x
du = 4 dx
dx = du / 4
I = (5/4) ∫ u^(-1/2).du
I = (5/4). u^(1/2) / (1/2) + C
I = (5/2).(2 + 4x)^(1/2) + C
I = (5/2).√(2 + 4x) + C