Your answer is not complete, as you need something after the last "plus". To give an answer that's still correct though, let's call it n:
If:
y = x2/(x2 + n)
dy/dx = [2x(x2 + n) - (2x + dn/dx) ] / (x2 + n)2
= (2x3 + 2nx - 2x - dn/dx) / (x4 + 2nx2 + n2)
= (2x3 + [2n - 2]x - dn/dx) / (x4 + 2nx2 + n2)
If n is a constant, then you can remove the term dn/dx, and simplify the term [2n - 2]x
14x
Your expression is this... x2 + y2/x
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
x + 5
x + 10
Negative the derivative of f(x), divided by f(x) squared. -f'(x) / f²(x)
8 + X SQUARED
14x
Your expression is this... x2 + y2/x
m
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
x + 3
x + 5
x + 10
x + 5
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
2