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 = 4u^1/2 + C = 4(x² + 3)^1/2 + C

2. dy = x² (2x³ + 5) ^1/2 dx

3. dy = x(2x² + 10)⁶ dx