# Integral of a Sqrt

#### PrudensOptimus

I try to take the integral of sqrt(4x-1) with respect to x...

The correct answer is sqrt((4x-1)^3)/6, but I always get

2sqrt((4x-1)^3)/3... Can someone explain how to solve that integral pls.

#### Tom Mattson

Staff Emeritus
Gold Member
Originally posted by PrudensOptimus
I try to take the integral of sqrt(4x-1) with respect to x...
OK

The correct answer is sqrt((4x-1)^3)/6,
No, it isn't.

Edit: My mistake; yes it is.

but I always get 2sqrt((4x-1)^3)/3...
That's not right either.

Can someone explain how to solve that integral pls.
Yes, you do a u-substitution. Let u=4x-1, so du=4dx. You then have:

(1/4)&int;u1/2du,

which is elementary.

Last edited:

#### JKLM

If you didn't understand what the last person said when you take the integral of squrt(4x-1) with respect to xdx you must first find u and u prime (or the derivative of u), where u is what is in the parentheses, in this case (4x-1) so u prime (or the derivative of (4x-1)) = 4x, but you still have to deal with the remaining x and the 4 left by the derivative. To get rid of the 4 in the derivative you divide the integral by 4, and subistute for the x so. x=(u-1)/4, I'm you understand the rest. You multiply u^(1/2)*((u-1)/4) then take the antiderivative of that.

#### HallsofIvy

The problem, by the way, is NOT the square root! You can easily integrate &int; &radic;(x) dx. &radic;(x)= x1/2 and you can use the "power rule".
The problem is that "4x-1" inside the square root. To get rid of that you make the substitution mentioned earlier: u= 4x-1 so that
&radic;(4x-1)= &radi;(u)= u1/2. Of course, you have to convert from "dx" to "du". Because 4x-1 is linear, that's easy
du/dx= 4 so du= 4 dx or (1/4)du= dx.

&int; &radic(4x-1)dx= (1/4)&int u1/2du.

The power rule says that an anti-derivative of un is
1/(n+1) un+1. In this case, n= 1/2 and n+1= 3/2. The anti-derivative is (2/3)u3/2+ C . Replacing u by 4x-1 again, &int; &radic(4x-1)dx= (2/3)(4x-1)3/2+ C.

Since the original problem was given in terms of &radic; rather than a 1/2 power, it might be a good idea to set the answer in those terms: &int; &radic(4x-1)dx= (2/3)(&radic;(4x-1))3+ C.

#### lavalamp

Why can't you just int. the bracket like you would int. x^0.5?

Code:
&int; (4x-1)^0.5 dx

increase the power and divied by it:

(4x-1)^1.5 + c = 2(&radic;(4x-1))^3 + c
----------       ------------
1.5                3

Last edited:

#### Sting

Why can't you just int. the bracket like you would int. x^0.5?
Because if you differentiate your result ((2/3)(4x-1)^(3/2) + C) you get:

4(4x - 1)(1/2)

which is not what you integrated.

#### Tom Mattson

Staff Emeritus
Gold Member
Originally posted by lavalamp
Why can't you just int. the bracket like you would int. x^0.5?

Code:
&int; (4x-1)^0.5 dx

increase the power and divied by it:

(4x-1)^1.5 + c = 2(&radic;(4x-1))^3 + c
----------       ------------
1.5                3
As Sting said, when you differentiate your result, you do not get the original integrand. That is because (4x-1)1/2 is a composition of functions.

f(u)=u1/2
u(x)=4x-1

In general, a composition of functions does not satisfy the same basic integration rule as the simple function f(u).

edit: typo

#### lavalamp

Sorry, I meant to integrate it as a sight integral, where a multiple of the differential of the bracket appears outside the bracket, ie:

&int; f'(x) * (f(x))^n dx

in this case f(x)=4x-1 and f'(x)=4. The mulitple in this case is 0.25 which you can take out of the integral and put it back in at the end to get the right answer:

Code:
1 * (4x-1)^1.5 + c = (&radic;(4x-1))^3 + c
-   ----------       ------------
4      1.5                6

#### synergy

Lavalamp,
That's how I learned to do it, way back in '87. I think it's much quicker and easier.
Aaron

#### lavalamp

I only learnt about integration at the end of 2002, and then I only learnt about sight integrals earlier this year. That's why I messed up the integration the first time round.

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