1. Integration problem

Hi, guys, I'm a newbie,

Could anyone help me please with this integral:

S ( (X^2) / sqrt (X^3 + 3) ) dx
(x squared over square root of ( x cubed plus three )

My steps are as follow:

U = ( x^3 + 3) du = 3x^2 dx;
dx = du / (3x^2) -- correct??
what next?

2. Originally Posted by Pupka
Hi, guys, I'm a newbie,

Could anyone help me please with this integral:

S ( (X^2) / sqrt (X^3 + 3) ) dx
(x squared over square root of ( x cubed plus three )

My steps are as follow:

U = ( x^3 + 3) du = 3x^2 dx;
dx = du / (3x^2) -- correct??
what next?
My approach to this type of integral is to recognise that 3 x^2 is the
derivative of x^3+3, so this suggest the integral is of the form:

integral (x^2) / sqrt (x^3 + 3) ) dx = k sqrt(x^3 + 3)

then we differentiate the right hand side and equate that to the integrand
to determine k.

(d/dx) k sqrt(x^3 + 3) = k (3x^2) (1/2) /sqrt(x^3+3)

......................=(3k/2) [x^2/sqrt(x^3+3)]

So we need 3k/2=1, or k=2/3, hence:

integral (x^2) / sqrt (x^3 + 3) ) dx = (2/3) sqrt(x^3 + 3)

RonL

3. Wow! Your reasoning is great! Thank you very much! I never used such method before... I mean where you found 2/3 using k. I'll keep in mind. I just read a chapter in my textbook on how to do this, and based on what I've understood, I solved the problem like this:

S ( (X^2) / sqrt (X^3 + 3) ) dx
u = (x^3 + 3)
dx = du / (3x^2)

so I have

S (x^2) / ( u )^(1/2) du / (3x^2) since my dx = du / (3x^2)

Now x^2 cancels in both numerator and denominator, 1/3 is taken up front and I am left with

1/3 S 1 / ( u )^1/2

= 1/3 S ( u )^-1/2, and integrating u gives us

1/3 * (u^1/2)/(1/2) =
1/3 * 2 *u^1/2 =
2/3 * sqrt (x^3 + 3).

Correct?