Calculus of a Single Variable, Chapter 6, 6.1, Section 6.1, Problem 49

The general solution of a differential equation in a form of y’ =f(x,y) can be 'evaluated using direct integration. The derivative of y denoted as y' can be written as(dy)/(dx) then y'= f(x,y) can be expressed as (dy)/(dx)= f(x,y) .
That is form of the given problem:(dy)/(dx)=xsqrt(x-6) .
We may apply the variable separable differential in which we follow N(y) dy = M(x) dx .
Cross-multiply dx to the right side: dy=xsqrt(x-6)dx .
Apply direct integration on both sides: intdy=int xsqrt(x-6)dx .
For the left side, we apply basic integration property:
int (dy)=y
For the right side, we may apply u-substitution by letting: u = x-6 or x = u+6 then dx = du .
intxsqrt(x-6)dx=int (u+6)sqrt(u)du
=int (u+6)u^(1/2)du
=int(u^(3/2)+6u^(1/2))du
Apply the basic integration property: int (u+v) dx= int (u) dx + int (v) dx .
int u^(3/2)du+ int 6u^(1/2)du
Apply the Power Rule for integration : int x^n= x^(n+1)/(n+1)+C .
int u^(3/2)du+ int 6u^(1/2)du=u^((3/2+1))/(3/2+1)+ 6u^((1/2+1))/(1/2+1)+C
=u^(5/2)/((5/2))+ 6u^(3/2)/((3/2))+C
=u^(5/2)*(2/5)+ 6u^(3/2)*(2/3)+C
=(2u^(5/2))/5+ 4u^(3/2)+C
Plug-in u = x-6 , we get:
intxsqrt(x-6)dx=(2(x-6)^(5/2))/5+ 4(x-6)^(3/2)+C
Combining the results, we get the general solution for the differential equation:
y=(2(x-6)^(5/2))/5+ 4(x-6)^(3/2)+C