Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 40

To evaluate the given integral problem: int_0^4 x/sqrt(3+2x)dx , we determine first the indefinite integral function F(x). From the table of indefinite integrals, we may consider the formula for integrals with roots as:
int u/sqrt(u+-a) du = 2/3(u-+2a)sqrt(u+-a)+C
Take note that we have "+ " sign inside the square root on int_0^4 x/sqrt(3+2x)dx then we will follow:
int u/sqrt(u+a) du = 2/3(u-2a)sqrt(u+a) +C.
We may let a = 3 and u = 2x or x= u/2
For the derivative of u, we get du = 2 dx or (du)/2 = dx .
Plug-in the values: u = 2x or x=u/2 ,and (du)/2 = dx , we get:
int_0^4 x/sqrt(3+2x)dx =int_0^4 (u/2)/sqrt(3+u)* (du)/2
=int_0^4 (u du)/(4sqrt(3+u))
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int_0^4 (u du)/(4sqrt(3+u)) =1/4 int_0^4 (u du)/sqrt(3+u)
Apply the aforementioned integral formula from the table of integrals, we get:
1/4 int_0^4 (u du)/sqrt(3+u) =1/4*[2/3(u-2(3))sqrt(u+3)]|_0^4
=1/4*[2/3(u-6)sqrt(u+3)]|_0^4
=2/12(u-6)sqrt(u+3)|_0^4
=1/6(u-6)sqrt(u+3)] |_0^4or((u-6)sqrt(u+3))/6|_0^4
Plug-in u = 2x on((u-6)sqrt(u+3))/6 +C , we get:
int_0^4 x/sqrt(3+2x)dx =((2x-6)sqrt(2x+3))/6|_0^4
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
((2x-6)sqrt(2x+3))/6|_0^4 =((2(4)-6)sqrt(2(4)+3))/6-((2(0)-6)sqrt(2(0)+3))/6
=((8-6)sqrt(8+3))/6- ((0-6)sqrt(0+3))/6
=(2*sqrt(11))/6- (-6sqrt(3))/6
= sqrt(11)/3-(-sqrt(3))
= sqrt(11)/3+sqrt(3)
= (sqrt(11)+3sqrt(3))/3 or 2.84 (approximated value).