Calculus of a Single Variable, Chapter 6, 6.1, Section 6.1, Problem 50

For the given problem:(dy)/(dx) =2xsqrt(4x^2+1) is a first order ordinary differential equation in a form of (dy)/(dx) = f(x,y) .
To evaluate this, we rearrange it in a form of variable separable differential equation: N(y) dy =M(x) dx .
Cross-multiply dx to the right side:dy=2xsqrt(4x^2+1)dx .
Apply direct integration on both sides: intdy= int 2xsqrt(4x^2+1)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 = 4x^2+1 then du=8x dx or (du)/8=x dx .
The integral becomes:
int 2xsqrt(4x^2+1)dx=int 2sqrt(u)*(du)/8
= int (sqrt(u)du)/4
We may apply the basic integration property: int c*f(x)dx= c int f(x) dx .
int (sqrt(u)du)/4= 1/4int sqrt(u)du
Apply Law of Exponent: sqrt(x)= x^(1/2) and Power Rule for integration : int x^n= x^(n+1)/(n+1)+C .
1/4int sqrt(u)du =(1/4) int u^(1/2)du
=(1/4)u^(1/2+1)/(1/2+1)+C
=(1/4)u^(3/2)/((3/2)) +C
=(1/4)u^(3/2)*(2/3) +C
=u^(3/2)/6+C
Plug-in u=4x^2+1 on u^(3/2)/6+C , we get:
int 2xsqrt(4x^2+1)dx=(4x^2+1)^(3/2)/6+C
Combining the results from both sides, we get the general solution of the differential equation as:
y=(4x^2+1)^(3/2)/6+C