Is this correct?

Evaluate 2x∫1 3t(t^2 + 1)^2 dt

u= t^2 + 1
du= 2t dt 1/2du=tdt
2x∫1 3u^2 1/2du
3/2 2x∫1 1/3u^3 + c
3/2 [1/3 u^3]
3/2(1/3(2x)^3) - 3/2(1/3(1)^3)
Answer= 4x - 1/2

To evaluate the definite integral ∫1 (3t(t^2 + 1)^2) dt, we can use the u-substitution method.

1. Start by setting u = t^2 + 1. This makes du = 2t dt. Rearranging, we have dt = du / (2t).

2. Substitute the new variables into the original integral:
∫1 (3t(t^2 + 1)^2) dt = ∫ (3(u - 1)^2) (du / (2t))

3. Simplify the integrand:
(3(u - 1)^2) / (2t) = (3/2) * (u - 1)^2 / t

4. The integral becomes:
∫ (3/2) * (u - 1)^2 / t du

5. Now, we can integrate with respect to u. Distribute (3/2) and split the integral into two parts:
(3/2) ∫ (u^2 - 2u + 1) / t du = (3/2) ∫ (u^2 / t - (2u / t) + (1 / t)) du

6. Integrate each term:
(3/2) * (u^3 / (3t) - 2(u^2 / (2t)) + (u / t))

7. Simplify:
(u^3 / (2t) - u^2 / t + u / t) + C

8. Substitute back u = t^2 + 1:
(t^3 + 1) / (2t) - (t^2 + 1) / t + (t^2 + 1) / t + C

9. Simplify further:
(t^3 + 1) / (2t) - (t^2 + 1) / t + 2 / t + C

10. Multiply through by 2t to clear the denominator:
(t^3 + 1) - 2(t^2 + 1) + 2 = t^3 - 2t^2 + 1

11. Finally, factor out a common term:
t^3 - 2t^2 + 1 = t^2(t - 2) + 1

So, the correct evaluation of the integral ∫1 (3t(t^2 + 1)^2) dt is:
t^2(t - 2) + 1 + C, where C is the constant of integration.