The solution of the differential equation $\dfrac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}$ is
A. $\left( {4x + y + 1} \right) = \tan \left( {2x + C} \right)$
B. ${\left( {4x + y + 1} \right)^2} = 2\tan \left( {2x + C} \right)$
C. ${\left( {4x + y + 1} \right)^3} = 3\tan \left( {2x + C} \right)$
D. $\left( {4x + y + 1} \right) = 2\tan \left( {2x + C} \right)$

Hint: Here, in the given question, we need to find the solution of the differential equation $\dfrac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}$. An equation containing an independent variable, dependent variable and differential coefficients of dependent variable w.r.t. an independent variable is called a differential equation. To find the solution of the given differential equation, we will use the substitution method. We will first put $4x + y + 1 = u$ and then we will differentiate this equation and simplify it. After this, we will integrate the equation to get the solution of the given differential equation.

Complete step by step solution:
 We have, $\dfrac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}$
Let $4x + y + 1 = u\,\,\,\,........\left( i \right)$
Now, differentiate $4x + y + 1 = u$ w.r.t. $x$
$ \Rightarrow 4 + \dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - 4$
Or,
$ \Rightarrow \dfrac{{du}}{{dx}} - 4 = \dfrac{{dy}}{{dx}}$
Given that, $\dfrac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}$. Therefore, we get
$ \Rightarrow \dfrac{{du}}{{dx}} - 4 = {\left( {4x + y + 1} \right)^2}$
From equation $\left( i \right)$, we get
$ \Rightarrow \dfrac{{du}}{{dx}} - 4 = {u^2}$
$ \Rightarrow \dfrac{{du}}{{dx}} = {u^2} + 4$
On cross multiplication, we get
$ \Rightarrow \dfrac{{du}}{{{u^2} + 4}} = dx$
Now, we will integrate the above written equation. We know that $\int {\dfrac{1}{{{a^2} + {x^2}}}} dx = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$. Therefore, we have
$ \Rightarrow \int {\dfrac{1}{{{2^2} + {u^2}}}du} = \int {1.dx} $
On integrating both the sides, we get
$ \Rightarrow \dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{u}{2}} \right) = x + C$
Now, substitute the value of $u$ from equation $\left( i \right)$
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{4x + y + 1}}{2}} \right) = 2x + C$
$ \Rightarrow \left( {\dfrac{{4x + y + 1}}{2}} \right) = \tan \left( {2x + C} \right)$
On cross multiplication, we get
$ \Rightarrow 4x + y + 1 = 2\tan \left( {2x + C} \right)$
Hence, the solution of the differential equation $\dfrac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}$ is $4x + y + 1 = 2\tan \left( {2x + C} \right)$

Option ‘D’ is correct

Note: To solve this type of questions, one must remember the rules of integration and differentiation. Always remember that the derivative of a constant with respect to a variable is always zero. We have an indefinite integral that is why we added integration constant. If we have a definite integral we do not add integration constant. Be careful while applying the integration formula as there is only a minor difference between the formulas.