Now look at this: dy = 2x(x2 + 1) 2dx
Remember: un = un+1/(n+1) + C
We can let u = x2 + 1. So du = 2x dx and n = 2.
Then we can substitute these into the first equation:
dy = u2 du
This is easy to integrate.
y = u3/3 + C
Then we plug the original stuff back in:
y = (x2+1) 2 + C
Try these and click on the number to see the answer. Integrate to find y. First do the substitution for u and du.
1. dy = 4x/(x2 + 3)1/2dx
u = x2 + 3 du = 2x dx n = -1/2
dy = 2u-1/2 du
y = 4u1/2 + C = 4(x2 + 3)1/2 + C
2. dy = x2 (2x3 + 5) 1/2 dx
3. dy = x(2x2 + 10)6 dx