Integration Continued

Integration continued…

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