Question

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

Answer 1

To find the antiderivative of $\dfrac{1}{1+{{x}^{2}}}$, we need to understand what an antiderivative means. Antiderivative means we have to do the integration of the given function. For that we are going to put x as $\tan \theta$ and then we will differentiate on both the sides of the equation. After that we will substitute x as $\tan \theta$ in the integration and also substitute $d\theta$ in place of dx. And hence, will do the integration.

Complete step by step solution:
In the above problem, we are asked to find the antiderivative of $\dfrac{1}{1+{{x}^{2}}}$. To find the antiderivative of $\dfrac{1}{1+{{x}^{2}}}$, we have to integrate this function $\dfrac{1}{1+{{x}^{2}}}$ with respect to x then we get,
$\int{\dfrac{1}{1+{{x}^{2}}}dx}$
Now, we are going to take $x=\tan \theta$ in the above integration and also, we are going to differentiate both the sides of this $x\And \theta$ relation.
$x=\tan \theta$
Differentiating both the sides of the above equation we get,
$dx={{\sec }^{2}}\theta d\theta$
Now, substituting the value of $x\And dx$ in the above integration we get,
$\int{\dfrac{1}{\left( 1+{{\tan }^{2}}\theta \right)}{{\sec }^{2}}\theta d\theta }$
We know the trigonometric identity which says that:
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$
Applying the above trigonometric identity in the above integration we get,
$\int{\dfrac{1}{\left( {{\sec }^{2}}\theta \right)}{{\sec }^{2}}\theta d\theta }$
As you can see that ${{\sec }^{2}}\theta$ is common in both the numerator and the denominator so it will be cancelled out and we are left with:
$\begin{aligned} & \int{d\theta } \\\ & =\theta +C \\\ \end{aligned}$
In the above expression, “C” is the constant.
Now, we are going to write $\theta$ in terms of x by using the following relation:
$x=\tan \theta$
Taking ${{\tan }^{-1}}$ on both the sides of the above equation we get,
${{\tan }^{-1}}x={{\tan }^{-1}}\tan \theta$
We know that if we multiply something with the inverse of something then we will get 1 so ${{\tan }^{-1}}\tan =1$ and using this relation in the above equation we get,
${{\tan }^{-1}}x=\theta$
Using the above relation in the above integration we get,
${{\tan }^{-1}}x+C$
Hence, the antiderivative of the above expression is equal to ${{\tan }^{-1}}x+C$.

Note: You can check the result of the antiderivative by taking the derivative of this integration and then see if you are getting this expression $\dfrac{1}{1+{{x}^{2}}}$ or not.
The result of the antiderivative of $\dfrac{1}{1+{{x}^{2}}}$ is equal to ${{\tan }^{-1}}x+C$.
Taking the derivative of ${{\tan }^{-1}}x+C$ with respect to x we get,
$\dfrac{d\left( {{\tan }^{-1}}x+C \right)}{dx}$
We know the differentiation of ${{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$ and the derivative of constant is 0 so using these derivatives we can find the above derivative as follows:
$\begin{aligned} & \dfrac{1}{1+{{x}^{2}}}+0 \\\ & =\dfrac{1}{1+{{x}^{2}}} \\\ \end{aligned}$
Hence, we are getting the same expression which we have started with.