Calculus question involving summation and multiplication series I was unable to solve this question properly, but as it was multiple choice, I scaled down the upper limit to $3$ and got an answer matching with the pattern. I am still unable to solve it.
I have tried to approach the question by trying to form a general term for integrand and then try to proceed using summation series.

$$
  \int_0^1
  \left( {} \sum_{k=1}^{2014} \frac{x^2}{x^3 + k^3} \right)
  \left( {} \prod_{k=1}^{2014} {} (x^3 + k^3) \right)
  \mathrm{d}x
  =
  \frac{1}{\lambda}
  \left[ {} \prod_{k=1}^{2014} {} (1 + k^3) - \mu^3 \right]
$$
Find $\lambda, \mu$.

 A: I can give you the general calculation and you can put in your number into any arbitrary arithmetic program. Swapping $2014$ for the general $N$, we have, following the first commenter's suggestion
$$\frac{d}{dx}\prod_{k=1}^{N} {} \left( x^{3} + k^{3}\right) = 3\sum_{k=1}^{N} \frac{x^{2}}{x^{3}+k^{3}}\prod_{k=1}^{N} {} \left( x^{3} + k^{3}\right)$$
This result is basically a fancy application of the ordinary product rule from calculus 1. If you want me to expand it, please ask, but my inclination is not to clutter this demonstration with material which is not very enlightening.
We therefore have
$$\int_0^1
  \left( {} \sum_{k=1}^{N} \frac{x^2}{x^3 + k^3} \right)
  \left( {} \prod_{k=1}^{N} {} (x^3 + k^3) \right)
  \mathrm{d}x$$
Substitute
$$u = \prod_{k=1}^{N} {} \left(x^3 + k^3\right)$$
then
$$\mathrm{d}u = 3\sum_{k=1}^{N} \frac{x^{2}}{x^{3}+k^{3}}\prod_{k=1}^{N} {} \left( x^{3} + k^{3}\right)\mathrm{d}x$$
and we now have
$$\frac{1}{3}\int_{x=0}^{x=1}\mathrm{d}u$$
where I used a little poor form and did not recalculate the limits of integration.
$$\left.\frac{1}{3}u\right|_{x=0}^{x=1}$$
$$\left.\frac{1}{3}\left(\prod_{k=1}^{N} {} (x^3 + k^3)\right)\right|_{x=0}^{x=1}$$
$$\frac{1}{3}\left(\prod_{k=1}^{N} {} (1^3 + k^3)\right)
- \frac{1}{3}\left(\prod_{k=1}^{N} {} (k^3)\right)$$
You now set this equal to the RHS of the problem
$$\frac{1}{3}\left(\prod_{k=1}^{N} {} (1 + k^3)\right)
- \frac{1}{3}\left(\prod_{k=1}^{N} {} (k^3)\right) = 
\frac{1}{\lambda}\left[ {} \prod_{k=1}^{N} {} (1 + k^3) - \mu^3 \right]$$
After rearranging I get
$$ \frac{\lambda}{3}\left(\prod_{k=1}^{N} {} (1 + k^3)
- \left(N!\right)^{3}\right) =
\prod_{k=1}^{N} {} (1 + k^3) - \mu^3$$
For you, $N = 2014$. Please let me know if there is a problem, but after setting N = 2014, you should have the relation between $\lambda$ and $\mu$. I don't see how there can be a unique answer. The "most obvious" answer seems to be to set $\lambda=3$ and $\mu=(N!)^3$, but I would think the relation has a family of answers.
Note, the commenter said to use $\frac{d}{du}\ln(u)=\frac{du}{dx}$, but I don't see how to do that. All I see is a $\mathrm{d}u$ form. Let me know.
Edit:
I just looked at it and this obviously cleans up to
$$\lambda = 3\frac{\prod_{k=1}^{N} {} (1 + k^3) - \mu^3}{\prod_{k=1}^{N} {} (1 + k^3) - \left(N!\right)^{3}}$$
which defines $\lambda$ as a function of $\mu$.
