I'll use ##c = 1##. Overdots mean proper-time derivatives.
Four-position: ## \mathbf{R} = (t, \mathbf{r}) ##
Four-velocity: ##\mathbf{V} = \dfrac{d \mathbf{R}}{d \tau} = (\gamma, \gamma \mathbf{v}) ##
Four-acceleration (requires some effort): ## \mathbf{A} = \dfrac{d \mathbf{V}}{d \tau} = (\gamma^4 (\mathbf{v} \cdot \mathbf{a}), \gamma^2 \mathbf{a} + \gamma^4 (\mathbf{v} \cdot \mathbf{a} ) \mathbf{v}) ##
In the case of rectilinear motion, ## \mathbf{v} \cdot \mathbf{a} = \pm va## and ##\mathbf{a} = \pm a \hat{\mathbf v}##, and the four-acceleration's components simplify to:
$$ \mathbf{A} = (\pm \gamma^4 v a, \gamma^4 \mathbf{a}) . $$
In this special case (that is, for all inertial observers for whom the accelerating object's motion is rectilinear), the magnitude of the four-acceleration can be reckoned from the components like so:
$$ \sqrt{ \mathbf{A} \cdot \mathbf{A} } = \sqrt{(\pm \gamma^4 v a)^2 - (\gamma^4 \mathbf{a})^2} = \gamma^3 a \, \mathrm{i} . $$
In the accelerating object's instantaneous rest frame, ##\gamma^3 = 1##, and the four-acceleration's magnitude (divided by ## \mathrm{i} ##) is just ##a##, which is the magnitude of acceleration that the object "feels." We'll call this the "proper acceleration" ##\alpha##; as the magnitude of a four-vector, its value is invariant, and any observer for whom the motion is rectilinear can calculate it like ##\alpha = \gamma^3 a##.
We'll come back to this result.
Because ##\gamma^2 - \gamma^2 v^2 = 1##, we can appeal to the identity ##\cosh^2 \phi - \sinh^2 \phi = 1## and introduce the "rapidity" ##\phi##:
## \cosh \phi = \gamma ,##
## \sinh \phi = \gamma v ,##
## \tanh \phi = \sinh \phi / \cosh \phi = v .##
This is especially useful for doing calculus, since, e.g., ##\sinh## and ##\cosh## are derivatives of each other.
Note in passing that ##\sinh## and ##\tanh## (and their inverses) are odd functions. That means we're free to define signed rapidities, too, which we can pair with (signed) velocity-components. I'll make some simplifications below to avoid dealing with them directly, but allowing for them isn't exactly rocket science. (Sorry.)
The four-velocity ##\mathbf{V} = (\gamma, \gamma \mathbf{v}) ## can now be rewritten:
$$ \mathbf{V} = (\cosh \phi, \hat{\mathbf v} \sinh \phi ) , $$
and four-acceleration (where the overdot means a proper-time derivative):
$$ \mathbf{A} = \dfrac{d \mathbf{V}}{d \tau} = (\dot{\phi} \sinh \phi, \hat{\mathbf v} \dot{\phi} \cosh \phi + \dot{\hat{\mathbf v}} \sinh \phi) . $$
Its magnitude is:
$$ A = \sqrt{ \mathbf{A} \cdot \mathbf{A} } = \sqrt{ (\dot{\phi} \sinh \phi)^2 - (\hat{\mathbf v} \dot{\phi} \cosh \phi + \dot{\hat{\mathbf v}} \sinh \phi)^2 } . $$
For rectilinear motion, ##\dot{\hat{\mathbf v}} = \mathbf 0##, and:
$$ A = \sqrt{ \dot{\phi}^2 ( \sinh^2 \phi - \cosh^2 \phi ) } = | \dot \phi | \, \mathrm{i} . $$
Put that together with our previous result for ##A## in the rectilinear case, and we have:
$$ \dfrac{d \phi}{d \tau} = \pm \alpha .$$
So for rectilinear motion, the rapidity's proper-time derivative is the proper acceleration, up to a sign (we defined ##\alpha## as positive).
Now we can integrate with respect to ##\tau## to get the rapidity as a function of proper time. (We'll actually get a signed rapidity in the general case, but we'll narrow things down in a moment.) If the proper acceleration is constant, that's:
$$ \int_0^\tau \dfrac{d \phi}{d \tau} d \tau = \pm \alpha \int_0^\tau d \tau , $$
giving ## \phi (\tau) = \phi_i \pm \alpha \tau ##, where ##\phi_i## is the initial rapidity (its value at ##\tau = 0##).
For convenience, let's now set the initial rapidity to zero and interest ourselves only in the "plus" case, so that our function is ## \phi (\tau) = \alpha \tau ##. By the definition of ##\phi##, we immediately also get:
##\gamma (\tau) = \cosh (\alpha \tau), ##
## (\gamma v)(\tau) = \sinh (\alpha \tau) ,##
## v(\tau) = \tanh (\alpha \tau ).##
How about position as a function of proper-time? Say the object accelerates in the positive ##x##-direction. Then we have ##\dot x = \gamma v_x = \gamma v##. To get ##x(\tau)##, integrate ##\gamma v = \sinh (\alpha \tau)##:
$$ \int_0^\tau \dfrac{d x}{d \tau} d \tau = \int_0^\tau \sinh(\alpha \tau) d \tau ,$$
giving ## x(\tau) = x_i + \cosh(\alpha \tau) / \alpha ##.
If you want functions of coordinate time ##t## instead, note that ##\alpha = \gamma^3 a = \frac{d (\gamma v)}{d t}## in the rectilinear case, integrate with respect to ##t## to get a function for ##\gamma v##, and use ##\phi = \sinh^{-1} (\gamma v)## to get ##\phi(t)##, etc.
Hope I didn't make any mistakes.
