The value of $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)$ is –
(a) $0$
(b) $1$
(c) $\dfrac{1}{2}$
(d) Non existent

Hint: The given expression is a combination of integration, differentiation and limit. So, first solve the innermost function and then proceed outwards. Here first solve the integration part, then apply the derivative and at last apply the limit.

Complete step-by-step answer:
 Given,
$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)$
First we will solve the integration part.
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{{{r}^{2}}-1}dr\]
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{r\left( {{r}^{2}} \right)}{{{r}^{2}}-1}dr\ldots \left( i \right)\]
Let,
\[u={{r}^{2}}-1\Rightarrow {{r}^{2}}=u+1\]
\[\Rightarrow \dfrac{du}{dr}=2r\]
\[\Rightarrow dr=\dfrac{1}{2r}du\]
Substituting these values in equation (i), we get
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{r\left( u+1 \right)}{u}\left( \dfrac{1}{2r}du \right)\]
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{\left( u+1 \right)}{u}\left( \dfrac{1}{2}du \right)\]
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{\left( u+1 \right)}{u}du\]
On expanding, we get
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\left( \dfrac{u}{u}+\dfrac{1}{u} \right)du\]
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\left( 1+\dfrac{1}{u} \right)du\]
This can be also written as,
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,1du+\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{1}{u}du \right]\]
But we know, $\mathop{\int }^{}\dfrac{1}{u}du=\ln u$ , so we get
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ u+\ln u \right]_{\sqrt{3}}^{\sqrt{x}}\]
Substituting back the values of ‘u’ and ‘du’, we get
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( {{r}^{2}}-1 \right)+\ln \left( {{r}^{2}}-1 \right) \right]_{\sqrt{3}}^{\sqrt{x}}\]
Applying the values, we get
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( {{\left( \sqrt{x} \right)}^{2}}-1 \right)+\ln \left( {{\left( \sqrt{x} \right)}^{2}}-1 \right) \right)-\left( \left( {{\left( \sqrt{3} \right)}^{2}}-1 \right)+\ln \left( {{\left( \sqrt{3} \right)}^{2}}-1 \right) \right) \right]\]
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( \left( 3-1 \right)+\ln \left( 3-1 \right) \right) \right]\]
\[\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( 2+\ln 2 \right) \right]\]
Now we will apply the derivative, we get
\[\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{d}{dx}\left\{ \dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( 2+\ln 2 \right) \right] \right\}\]
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{d}{dx}\left[ \left( x-1 \right)+\ln \left( x-1 \right) \right]-\dfrac{d}{dx}\left[ 2+\ln 2 \right] \right\}\]
But differentiation of constant is a constant, so
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{d}{dx}\left[ \left( x-1 \right)+\ln \left( x-1 \right) \right]-0 \right\}\]
This can be written as,
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{d}{dx}\left[ x \right]-\dfrac{d}{dx}\left[ 1 \right]+\dfrac{d}{dx}\left[ \ln \left( x-1 \right) \right] \right\}\]
Solving we get,
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ 1-0+\dfrac{d}{dx}\left[ \ln \left( x-1 \right) \right] \right\}\]
Differentiation of $\ln y=\dfrac{1}{y}.\dfrac{dy}{dx}$ , so we get
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ 1+\dfrac{1}{x-1}.\dfrac{d}{dx}\left[ x-1 \right] \right\}\]
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ 1+\dfrac{1}{x-1}.\left( 1 \right) \right\}\]
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{\left( x-1 \right)+1}{x-1} \right\}\]
\[\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{x}{x-1} \right\}\]
Now we will apply limit, we get
\[\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\mathop{\overset{\sqrt{x}}{\mathop{\int }}\,}}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\{ \dfrac{x}{x-1} \right\}\]
\[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\{ \dfrac{1}{\dfrac{\left( x-1 \right)}{x}} \right\}\]
\[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\{ \dfrac{1}{1-\dfrac{1}{x}} \right\}\]
Now we will apply the limit, we get
\[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{1}{1-\dfrac{1}{\infty }} \right\}\]
We know, $\dfrac{1}{\infty }\approx 0$ ,
\[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\{ \dfrac{1}{1-0} \right\}\]
\[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\]
Hence, the correct option for the given question is option (c).
Answer is Option (c)

Note: In the solution the following equation,
\[\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr\]
It can be solved by applying a partial fraction decomposition method too. The answer won’t differ.