How do you integrate \[\dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\]?

Hint: Here we will integrate the given equation by substituting the value of \[x\] as \[\tan \theta \]. Then we will differentiate the assume function and substitute the values in the given integrand. We will simplify the equation using trigonometric identities and formulas. We will then use integration formula to perform integration. After that we will put the value of \[\theta \] in the equation to get the integration in terms of \[x\].

Complete step-by-step answer:
Given equation is \[\dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\].
Let \[I\] be the integration of the given equation. Therefore we can write the integration equation as
\[I = \int {\dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}dx} \]……………\[\left( 1 \right)\]
Now we will put the value of \[x\] as \[\tan \theta \] i.e.
\[x = \tan \theta \]
So by differentiating it, we get
\[dx = {\sec ^2}\theta d\theta \]
Therefore, by putting these values in the equation \[\left( 1 \right)\], we get
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \]
We know that \[1 + {\tan ^2}\theta = {\sec ^2}\theta \].
Replacing \[1 + {\tan ^2}\theta \] by \[{\sec ^2}\theta \] in the above equation, we get
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^4}\theta }}} \]
Dividing the numerator and denominator by \[{\sec ^2}\theta \],we get
\[ \Rightarrow I = \int {\dfrac{{d\theta }}{{{{\sec }^2}\theta }}} \]
We know that the value of \[\sec \theta \] is equal to \[\dfrac{1}{{\cos \theta }}\] i.e. \[\sec \theta = \dfrac{1}{{\cos \theta }}\]. Therefore putting this value in the above equation, we get
\[ \Rightarrow I = \int {{{\cos }^2}\theta d\theta } \]
We know that \[\cos 2\theta = 2{\cos ^2}\theta - 1\]. Therefore, using this trigonometric property we will get the value of \[{\cos ^2}\theta \] as \[\dfrac{{\cos 2\theta + 1}}{2}\]. Therefore by putting this value, we get
\[ \Rightarrow I = \int {\dfrac{{\cos 2\theta + 1}}{2}d\theta } \]
\[ \Rightarrow I = \dfrac{1}{2}\int {\left( {\cos 2\theta + 1} \right)d\theta } \]
Integrating the terms, we get
\[ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{{\sin 2\theta }}{2} + \theta } \right) + C\]
\[ \Rightarrow I = \dfrac{{\sin 2\theta }}{4} + \dfrac{1}{2}\theta + C\]
We know that \[\sin 2\theta = 2\sin \theta \cos \theta \]. Therefore putting this value in the above equation, we get
\[ \Rightarrow I = \dfrac{{2\sin \theta \cos \theta }}{4} + \dfrac{1}{2}\theta + C\]
\[ \Rightarrow I = \dfrac{1}{2}\sin \theta \cos \theta + \dfrac{1}{2}\theta + C\]
Now we will put the value of \[\theta \] in the equation to get the equation in terms of \[x\]. So, we will calculate the value of \[\theta \] accordingly.
We know that \[x = \tan \theta \] we can write it as \[\tan \theta = \dfrac{x}{1}\]. So by this we can conclude that this is for a right triangle with side opposite to \[\theta \] is equal to \[x\] and side adjacent to it is equal to 1. Then by using the Pythagoras theorem we will get the hypotenuse which will be equal to \[\sqrt {1 + {x^2}} \]. Therefore the value of \[\sin \theta \] and \[\cos \theta \] is
\[\sin \theta = \dfrac{x}{{\sqrt {1 + {x^2}} }}\] and \[\cos \theta = \dfrac{1}{{\sqrt {1 + {x^2}} }}\]
So by using this we will put the value of \[\theta \] in the integral equation. Therefore, we get
\[ \Rightarrow I = \dfrac{1}{2}\sin \left( {{{\sin }^{ - 1}}\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)\cos \left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right) + \dfrac{1}{2}{\tan ^{ - 1}}x + C\]
Now by solving this we get
\[ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right) + \dfrac{1}{2}{\tan ^{ - 1}}x + C\]
\[ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{x}{{1 + {x^2}}}} \right) + \dfrac{1}{2}{\tan ^{ - 1}}x + C\]
Hence, the integration of \[\dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\] is equal to \[\dfrac{1}{2}\left( {\dfrac{x}{{1 + {x^2}}}} \right) + \dfrac{1}{2}{\tan ^{ - 1}}x + C\].

Note: Integration is defined as the process of summation of all the discrete data. In order to solve this question we should know the basic formula of the integration by parts of an equation. We have to remember to put the constant term \[C\] after the integration of an equation. We will use the basic trigonometric functions and Pythagoras theorem to get the value of \[\theta \] in terms of the inverse function of sin and cos to get the integration in the simplified form and in terms of \[x\].