Question

What is the integral of `int 1/(x^2+4)dx`?

Answer 1

`1/2tan^-1(x/2)+C`

Explanation:

The integral of `1/(x^2+a^2)` is given by:
`1/atan^-1(x/a)`

In our case, `a=2` (because `a^2=(2)^2=4`) and `x=x`, of course. Making necessary substitutions into the expression above, we have:
`int 1/(x^2+4)dx=1/2tan^-1(x/2)+C` (never forget the constant `C`)

If you want to see why, I'll describe the process below. I apologize if it gets a bit long, but sometimes calculus can be like that. Feel free to skip over it if it bores you.

First, we note that our denominator, `x^2+a^2`, looks like the Pythagorean theorem: `a^2+b^2=c^2`, the famous equation that describes the relationship between the legs of a right triangle. With that in mind, we can safely assume that `x` and `a` can actually be represented by a right triangle, as shown in the picture below:

We can see from this picture that the tangent of angle `theta` is equal `x/a`, or `tan(theta)=x/a`. (We use tangent and not sine or cosine because tangent includes both `x` and `a`, and the `x` is in the numerator which makes it look nicer).

If `tan(theta)=x/a`, then by multiplying both sides by `a` we get
`atan(theta)=x`. We can now substitute this expression into the integral to solve it. However, if we replace `x` with `theta`, we must also replace `dx` with `d theta`. We do this by differentiating `atan(theta)=x` with respect to `theta`:
`dx/(d theta)=asec^2(theta)` (because the derivative of `tan(theta)` is `sec^2(theta))`
`dx=asec^2(theta)d theta`

We are now ready to make all substitutions:
`int 1/(x^2+a^2)dx=int 1/(((atan(theta))^2+a^2))*asec^2(theta)d theta`

Now with some simplifying:
`int 1/(x^2+a^2)dx=int (asec^2(theta))/(a^2tan^2(theta)+a^2)d theta`
`int 1/(x^2+a^2)dx=int (asec^2(theta))/(a^2(tan^2(theta)+1))d theta`
`int 1/(x^2+a^2)dx=a/a^2int sec^2(theta)/(tan^2(theta)+1)d theta`
`int 1/(x^2+a^2)dx=1/aint sec^2(theta)/(tan^2(theta)+1)d theta`

If we remember our Pythagorean identities, we know that `tan^2(theta)+1=sec^2(theta)`. Because we have `tan^2(theta)+1` (in the denominator), we can replace it with `sec^2(theta)`:
`int 1/(x^2+a^2)dx=1/aint sec^2(theta)/(sec^2(theta))d theta`
`int 1/(x^2+a^2)dx=1/aint 1d theta`

Our result is `1/a*theta`. However, we need to convert this result back to `x`. How? Earlier, we said `atan(theta)=x`. That means `tan(theta)=x/a`, and furthermore, `theta=tan^-1(x/a)`. Plugging this into our result, we have
`int 1/(x^2+a^2)dx=1/atan^-1(x/a)`

And voila, there you have it.