Question regarding tangent and fraction integration

Question

I am studying integration and differentiation again and have come across an equation that I can't wrap my head around.

$$\int{\frac{1}{x^2 + a^2}}dx = \frac{1}{a}\tan^{-1}\Bigr(\frac{x}{a}\Bigr) + C$$

This is one of the basic integration equations regarding fractions, but I'm having trouble understanding how it came to be, and I don't want to just memorize it. Could someone be kind enough explain how the right hand side was derived?

Thank you.

Answer 1

Applying the formula for the derivative of an inverse function, we know that for any $x\in \mathbb{R}$, $$arctan'(x)=\frac{1}{tan'(arctan(x))}$$

Now, recall that for any $\theta \in \mathbb{R}$, $\theta \notin \pi\mathbb{Z}+\frac{\pi}{2}$, we have $$tan'(\theta)=1+tan(\theta)^2$$

Hence, $$arctan'(x)=\frac{1}{1+x^2}$$

It then follows that $$arctan'\left(\frac{x}{a}\right)=\frac{1}{a}\times\frac{1}{1+\left(\frac{x}{a}\right)^2}=\frac{a}{a^2+x^2}$$

Answer 2

Let $u = \dfrac xa, a \ne 0\implies\,\mathrm dx = a\,\mathrm du$. $$\begin{align} \int\dfrac1{x^2 + a^2}\,\mathrm dx &= \int\dfrac{a}{a^2u^2+a^2}\,\mathrm du \\ &= \dfrac1a\int\dfrac1{u^2 + 1}\,\mathrm du \\ &= \dfrac1a\arctan(u) + C \\ &= \dfrac1a\arctan\left(\dfrac xa\right) + C \end{align}$$

Answer 3

There are many ways to go about this question: you could first factor out an $a^2$ from the denominator and then use a convenient u-substitution on $u=x/a$, or you can let $x=a u$ and then factor out the $a^2$.

Let $x=au$ so $dx=a\, du$. Therefore, the integral becomes$$\begin{align*}I & =a\int du\frac 1{(au)^2+a^2}\\ & =\frac 1a\int du\,\frac 1{u^2+1}\\ & =\frac 1a\arctan u+C\\ & =\frac 1a\arctan\frac xa+C\end{align*}$$

An alternative would be to substitute $x=a\tan u$ and note that$$\frac {dx}{1+x^2}=\frac {du}a$$

Therefore, the transformation turns the integral into something much simpler to d eal with.