Question

Find the integral of the function $\frac{1}{x^2+a^2}$

• A. ${\tan }^{-1}\left(\frac{x}{a}\right)$
• B. $\frac{1}{a}{\tan }^{-1}\left(x\right)$
• C. $\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$
• D. ${\tan }^{-1}\left(\frac{x}{a}\right)$

Answer 1

The correct option is C $\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$

This is one of the standard integrals related to inverse trigonometric functions. Try to guess an inverse trigonometric function whose derivative is similar to the function $\frac{1}{x^2+a^2}$. It would be ${\tan }^{-1}\left(x\right)$. We know that the derivative of ${\tan }^{-1}\left(x\right)$ is $\frac{1}{x^2+1}$. This is different from the derivative given to us. To convert it into an integrable form, we will take $a^2$ from the denominator.
So we get $\frac{1}{a^2}\frac{1}{{\left(\frac{x}{a}\right)}^2+1}$
This could be related to the derivative of ${\tan }^{-1}\left(\frac{x}{a}\right)$, because instead of x, we have $\frac{x}{a}$.
To check that, let’s find the derivative of ${\tan }^{-1}\left(\frac{x}{a}\right)$ It would be $\frac{a}{x^2+a^2}$.
$\Rightarrow \frac{d}{dx}{\tan }^{-1}\left(\frac{x}{a}\right)=\frac{a}{x^2+a^2}\Rightarrow \int \frac{a}{x^2+a^2}dx={\tan }^{-1}\left(\frac{x}{a}\right)\Rightarrow \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$