If $P=\displaystyle \lim_{n\to \infty }\dfrac{{{\left( \prod\limits_{r=1}^{n}{\left( {{n}^{3}}+{{r}^{3}} \right)} \right)}^{\dfrac{1}{n}}}}{{{n}^{3}}}$ and $\lambda =\int_{0}^{1}{\dfrac{1}{{{x}^{3}}+1}}$, then lnP is equal to
[a] $\ln 2-1+\lambda $
[b] $\ln 2-3+3\lambda $
[c] $2\ln 2-\lambda $
[d] $\ln 4-3+4\lambda $

Hint: In the expression for P take log on both sides and use the fact that $\log \left( \prod\limits_{r=1}^{n}{f\left( r \right)} \right)=\sum\limits_{r=1}^{n}{\log \left( f\left( r \right) \right)}$ and $\log \left( {{a}^{n}} \right)=n\log a$. Hence prove that $\log P=\dfrac{1}{n}\sum\limits_{r=1}^{n}{\log \left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)}$. Use the fact that $\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\sum\limits_{r=x}^{y}{f\left( \dfrac{r}{n} \right)=\int_{a}^{b}{f\left( x \right)dx}}$ where $a=\displaystyle \lim_{n\to \infty }\dfrac{x}{n}$ and $b=\displaystyle \lim_{n\to \infty }\dfrac{y}{n}$. Hence prove that $\ln P=\int_{0}^{1}{\log \left( 1+{{x}^{3}} \right)dx}$ Using integration by parts express $\ln P$ in terms of $\lambda $ and hence find which of the options is correct.

Complete step-by-step answer:
We have
$P=\displaystyle \lim_{n\to \infty }\dfrac{{{\left( \prod\limits_{r=1}^{n}{\left( {{n}^{3}}+{{r}^{3}} \right)} \right)}^{\dfrac{1}{n}}}}{{{n}^{3}}}$
Hence, we have
$P=\displaystyle \lim_{n\to \infty }\dfrac{{{\left( \prod\limits_{r=1}^{n}{{{n}^{3}}\left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)} \right)}^{\dfrac{1}{n}}}}{{{n}^{3}}}$
We know that $\prod\limits_{r=1}^{n}{af\left( r \right)}={{a}^{n}}\prod\limits_{r=1}^{n}{f\left( r \right)}$, where a is independent of r.
Hence, we have
$P=\displaystyle \lim_{n\to \infty }\dfrac{{{\left( {{n}^{3n}}\prod\limits_{r=1}^{n}{\left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)} \right)}^{\dfrac{1}{n}}}}{{{n}^{3}}}=\displaystyle \lim_{n\to \infty }{{\left( \prod\limits_{r=1}^{n}{\left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)} \right)}^{\dfrac{1}{n}}}$
Taking log on both sides, we get
$\log P=\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\log \left( \prod\limits_{r=1}^{n}{\left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)} \right)$
We know that $\log \left( \prod\limits_{r=1}^{n}{f\left( r \right)} \right)=\sum\limits_{r=1}^{n}{\log \left( f\left( r \right) \right)}$
Hence, we have
$\log P=\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\sum\limits_{r=1}^{n}{\log \left( 1+\dfrac{{{r}^{3}}}{{{n}^{3}}} \right)}$
We know that $\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\sum\limits_{r=x}^{y}{f\left( \dfrac{r}{n} \right)=\int_{a}^{b}{f\left( x \right)dx}}$ where $a=\displaystyle \lim_{n\to \infty }\dfrac{x}{n}$ and $b=\displaystyle \lim_{n\to \infty }\dfrac{y}{n}$.
Hence, we have
$\log P=\int_{0}^{1}{\log \left( 1+{{x}^{3}} \right)dx}$
We know that if $\int{f\left( x \right)dx}=u\left( x \right)$ and $\dfrac{d}{dx}g\left( x \right)=v\left( x \right)$, then $\int{f\left( x \right)g\left( x \right)dx}=g\left( x \right)u\left( x \right)-\int{u\left( x \right)v\left( x \right)dx}$. This is known as integration by parts.
The function f(x) is called the second function and the function g(x) is called the first function.
The order of preference (in general) for choosing first function is given by ILATE rule
I = Inverse Trigonometric
L = Logarithmic
A = Algebraic
T = Trigonometric
E = Exponential
Using the above rule, we will take f(x) =1 and $g\left( x \right)=\log \left( 1+{{x}^{3}} \right)$, and we have
$u\left( x \right)=\int{1dx}=x$ and $v\left( x \right)=\dfrac{d}{dx}\left( \log \left( 1+{{x}^{3}} \right) \right)=\dfrac{3{{x}^{2}}}{1+{{x}^{3}}}$
Hence, we have
$\begin{align}
  & \log P=\left. \left( \log \left( 1+{{x}^{3}} \right)x \right) \right|_{0}^{1}-\int_{0}^{1}{\dfrac{3{{x}^{3}}}{{{x}^{3}}+1}dx} \\
 & =\ln 2-3\int_{0}^{1}{\dfrac{{{x}^{3}}}{{{x}^{3}}+1}dx} \\
\end{align}$
Adding and subtracting 1 in the numerator of the integrand, we get
$\begin{align}
  & \log P=\ln 2-3\int_{0}^{1}{\dfrac{{{x}^{3}}+1-1}{{{x}^{3}}+1}dx} \\
 & =\ln 2-3\int_{0}^{1}{1dx}+3\int_{0}^{1}{\dfrac{dx}{{{x}^{3}}+1}} \\
\end{align}$
Hence, we have
$\log P=\ln 2-3+3\lambda $

So, the correct answer is “Option b”.

Note: [1] The formula $\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\sum\limits_{r=x}^{y}{f\left( \dfrac{r}{n} \right)=\int_{a}^{b}{f\left( x \right)dx}}$ where $a=\displaystyle \lim_{n\to \infty }\dfrac{x}{n}$ and $b=\displaystyle \lim_{n\to \infty }\dfrac{y}{n}$ is actually definition of definite integral as a limit of a sum. We can memorise the formula by keeping the following relations in mind
$\sum{{}}\to \int{{}},x \to a,y\to b,\dfrac{r}{n}\to x,\dfrac{1}{n}\to dx$