`int_0^4 x/sqrt(1 + 2x) dx` Evaluate the definite integral.

Make the substitution `1+2x=u,` then `x=(u-1)/2` and `dx=(du)/2.`

The indefinite integral is

`int (1/2*(u-1)/2)/(sqrt(u)) du=(1/4)*int(sqrt(u)-1/sqrt(u)) du=(1/4)*((2/3)u^(3/2)-2u^(1/2))+C=`

`=(1/2)*(1/3)*sqrt(u)*(u-3)+C=(1/6)*sqrt(1+2x)*(2x-2)+C=`

`=(1/3)(x-1)sqrt(1+2x)+C.`

Therefore the definite integral is

`(1/3)(3*sqrt(9)-(-1))=10/3` =3 and 1/3 =3.(3).