`int_0^4 x/sqrt(3+2x) dx` Use integration tables to evaluate the definite integral.

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 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`


                ` =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).

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