Given the simple first order reaction $\ce{A -> P}$ 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?