parting integral into two parts, is this correct? I have: 
$$ \int_2^5 \mathrm{\frac{x-2}{x}}\,\mathrm{d}x $$
I tried: 
$$ \int_2^5 \mathrm{\frac{x-2}{x}}\,\mathrm{d}x = \int_2^5 \mathrm{(x-2)}\,\mathrm{d}x \cdot \int_2^5 \mathrm{\frac{1}{x}}\,\mathrm{d}x = [\frac{x^2}{2}+2x]_2^5 \cdot [ln(x)]_2^5 $$
is this correct? i have the feeling, i am messing up the borders of integrals..
 A: No. Instead you can write the integrand as:
$$\frac{x-2}{x}=\frac{x}{x}-\frac{2}{x}=1-2\cdot\frac{1}{x}$$
A: Yes, sorry to say that, but you misunderstood some integral properties...First of all $$\int_2^5 \mathrm{\frac{x-2}{x}}\,\mathrm{d}x = \int_2^5 1-\frac2x\,\mathrm{d}x=\int_2^5 \,\mathrm{d}x-2\int_2^5 \frac1x\,\mathrm{d}x=[x]_2^5-2[\ln x]_2^5=5-2-2(\ln 5 -\ln 2)=3-2\ln\left(\frac52\right)$$
Your method is wrong, as it is not always true that $$\int f(x)\cdot g(x) \,\mathrm{d}x = \int f(x) \,\mathrm{d}x \cdot \int g(x) \,\mathrm{d}x$$
A: There's no hope that $\int f(x)g(x)~\mathrm{d}x=\int f(x)~\mathrm{d}x\int g(x)~\mathrm{d}x$ will lead you to the correct answer in general. Instead, you can write your fraction as: $$\dfrac{x-2}{x}=\dfrac{x}{x}-\dfrac{2}x=1-\dfrac2x.$$
Now, and since integrals are linear, you have: $$\int\dfrac{x-2}{x}\mathrm{d}x=\int\left[1-\dfrac2x\right]\mathrm{d}x=\int1~\mathrm{d}x+\int-\dfrac2x\mathrm{d}x.$$
The last integral can be evaluated by recalling that: $$\int\dfrac1x\mathrm dx=\ln|x|+\rm {const}.$$
For a bit more complicated fractions like $\tfrac{4x^3-x+1}{-x^3+4x}$ you can use the more general technique of decomposing fractions.
A: No.  It is not true in general that $$\int f(x) g(x) \, dx = \int f(x) \, dx \int g(x) \, dx.$$
