Integrated rate law graphs
The short answer is that the equations you describe as graphs are the solutions of the differential equations that define the rate laws in question.  If you don't understand what a differential equation is, any sensible answer is going to be problematic.

Assuming you do know what a differential equation is, we proceed!

For a first order rate law, the differential equation is something like:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$

You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:

$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

Similarly, the differential equation for the second order rate law is:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$

And again integrate to get:

$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

This all comes from the following Wikipedia page, which has a lot more details: Rate equation