Integration of the first order rate equation Given the simple first order reaction $\ce{A -&gt; P}$ derive the integrated rate law. 

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

Now for the bit I need help with, the integration: 
Apparently the integrated form of $\frac{\mathrm d[\ce{A}]}{[\ce{A}]}= \ln[\ce{A}]$ but I'm struggling to see exactly how. 

The next step (usually omitted)I think should be to split the fraction before integration
$$\frac{\mathrm d[\ce{A}]}{1}\frac{1}{[\ce{A}]}=\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}=-kt$$

$$\int{\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}}=\int{-kt}$$

$$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=\int{-kt}$$

Now, to evaluate this I use the general integration law for a constant $\int{[\ce{A}]\,\mathrm dt} = [\ce{A}]t + C$ and reciprocal  $\int{\frac{1}{[\ce{A}]}=\ln[\ce{A}]+C}$. Therefore $$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=[\ce{A}]t + C_1+ \ln[\ce{A}]+C_2=\int{\frac{\mathrm d[\ce{A}]}{[\ce{A}]}}$$

I know this is wrong because the textbooks say something different but I don't know how to correct it. Can anyone help?
Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.

The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$

Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$

Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$

Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$

Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$

Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$

Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$