Question

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

Answer 1

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
$\begin{aligned} & \Rightarrow \dfrac{dx}{dx}={{\sec }^{2}}\theta \dfrac{d\theta }{dx} \\\ & \Rightarrow 1={{\sec }^{2}}\theta \dfrac{d\theta }{dx} \\\ & \Rightarrow {{\sec }^{2}}\theta \dfrac{d\theta }{dx}=1 \\\ \end{aligned}$
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

& \Rightarrow I=\int{\dfrac{{{\sec }^{2}}\theta d\theta }{{{\sec }^{2}}\theta }} \\\ & \Rightarrow I=\int{d\theta } \\\ \end{aligned}$$ On integrating the above expression, we get $\Rightarrow I=\theta +C........\left( iv \right)$ From the equation (ii) we get $\Rightarrow x=\tan \theta $ Taking inverse tangent both the sides, we get $$\begin{aligned} & \Rightarrow {{\tan }^{-1}}x={{\tan }^{-1}}\left( \tan \theta \right) \\\ & \Rightarrow {{\tan }^{-1}}x=\theta \\\ & \Rightarrow \theta ={{\tan }^{-1}}x \\\ \end{aligned}$$ Finally, substituting the above equation in the equation (iv) we get $\Rightarrow I={{\tan }^{-1}}x+C$ **Hence, we have finally integrated the given expression and obtained the integral equal to $${{\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.