Question

What‌ ‌is‌ ‌the‌ ‌antiderivative‌ ‌of‌ ‌$\dfrac{1}{{{x}^{2}}+4}$‌ ‌?‌ ‌

Answer 1

Hint: The antiderivative of a function can be found out by integrating it. The integration can be done by comparing the given function to some known derivative. Once this is done we can find the antiderivative of the given function.

Complete step by step solution:
The given function is $\dfrac{1}{{{x}^{2}}+4}$ . We need to find its antiderivative. We know that the derivative of an inverse tangent function is given by,
$\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{{{x}^{2}}+1}$
Here, replacing x by $\dfrac{x}{a}$ , we get,
$\dfrac{d\left( {{\tan }^{-1}}\left( \dfrac{x}{a} \right) \right)}{dx}=\dfrac{a}{{{x}^{2}}+{{a}^{2}}}$
By rearranging, we get
$\dfrac{1}{a}\dfrac{d\left( {{\tan }^{-1}}\left( \dfrac{x}{a} \right) \right)}{dx}=\dfrac{1}{{{x}^{2}}+{{a}^{2}}} \\ $
We know that antiderivative means integration, hence on taking integral both sides
$\Rightarrow \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=}\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+c \\ $
Hence we have obtained a generalized antiderivative for a function of the form $\dfrac{1}{{{x}^{2}}+{{a}^{2}}}$. Comparing it to the function given to us in the question, we can clearly see that,
$\begin{align}
  & {{a}^{2}}=4 \\
 & \,\,a=2 \\
\end{align}$
Therefore, now we are in a position to find the antiderivative for the function given, we get,
$\int{\dfrac{1}{{{x}^{2}}+{{2}^{2}}}dx=}\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+c$ , where c is the constant of indefinite integration.
Hence, the antiderivative of $\dfrac{1}{{{x}^{2}}+4}$is $\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+c$.

Note: While finding the antiderivative, one has to be aware of the chain rule which is applied during differentiation. Care has to be taken to make sure that the antiderivative takes into account the chain rule applied on the original function. For example, in the above question, the chain rule came into the picture as we were differentiating ${{\tan }^{-1}}\left( \dfrac{x}{a} \right)$ and hence, we had to take $\dfrac{1}{a}$ in the antiderivative as a direct consequence of the chain rule of differentiation.