Question

What is the integral of $\dfrac{1}{1+{{x}^{2}}}$?

Answer 1

Hint: In the question we are given a fractional expression to integrate. It cannot be integrated directly. For integrating it, we need to use the trigonometric substitution $x=\tan \theta $ into the given expression. After substitution, the given expression will become $\dfrac{1}{1+{{\tan }^{2}}\theta }$, which can be simplified using the trigonometric identity given by $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $. Also, we have to differentiate the equation $x=\tan \theta $ to get the value \[dx={{\sec }^{2}}\theta d\theta \] so that the given integral will get transformed from $x$ to $\theta $. The integral obtained in terms of \[\theta \] can be easily integrated and finally substituting $\theta ={{\tan }^{-1}}x$ we will obtain the final integrated expression.

Complete step by step solution:
According to the question, we have to integrate the function $\dfrac{1}{1+{{x}^{2}}}$. So we can write the integral as
$\Rightarrow I=\int{\dfrac{dx}{1+{{x}^{2}}}}......\left( i \right)$
For integrating the given expression, let us substitute
$\Rightarrow x=\tan \theta .......\left( ii \right)$
Differentiating both the sides with respect to x, we get
$\begin{align}
  & \Rightarrow \dfrac{dx}{dx}={{\sec }^{2}}\theta \dfrac{d\theta }{dx} \\
 & \Rightarrow 1={{\sec }^{2}}\theta \dfrac{d\theta }{dx} \\
 & \Rightarrow {{\sec }^{2}}\theta \dfrac{d\theta }{dx}=1 \\
\end{align}$
Multiplying the above equation by $dx$ we get
$\Rightarrow {{\sec }^{2}}\theta d\theta =dx.......\left( iii \right)$
Substituting the equations (ii) and (iii) into the equation (i) we get the integral as
\[\Rightarrow I=\int{\dfrac{{{\sec }^{2}}\theta d\theta }{1+{{\tan }^{2}}\theta }}\]
Now, we know the trigonometric identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $. On substituting this in the above expression, we get
\[\begin{align}
  & \Rightarrow I=\int{\dfrac{{{\sec }^{2}}\theta d\theta }{{{\sec }^{2}}\theta }} \\
 & \Rightarrow I=\int{d\theta } \\
\end{align}\]
On integrating the above expression, we get
$\Rightarrow I=\theta +C........\left( iv \right)$
From the equation (ii) we get
$\Rightarrow x=\tan \theta $
Taking inverse tangent both the sides, we get
\[\begin{align}
  & \Rightarrow {{\tan }^{-1}}x={{\tan }^{-1}}\left( \tan \theta \right) \\
 & \Rightarrow {{\tan }^{-1}}x=\theta \\
 & \Rightarrow \theta ={{\tan }^{-1}}x \\
\end{align}\]
Finally, substituting the above equation in the equation (iv) we get
$\Rightarrow I={{\tan }^{-1}}x+C$
Hence, we have finally integrated the given expression and obtained the integral equal to \[{{\tan }^{-1}}x+C\], where C is a constant.

Note: The integral given in the question was an indefinite integral. So do not forget to add a constant after integrating. Also, do not forget to obtain $dx$ in the form of $d\theta $ since there must be only a single variable in an integral.