Evaluate the following integral, if possible: $$\int_1^4 \frac{w}{w-3}dw$$
$$u= w-3,\; du = dw$$ $$\int_1^4 \frac{u+3}{u}du \Rightarrow \int_1^4 \frac{u}{u}du + \int_1^4 \frac{3}{u}du$$ $$\Rightarrow w-3 \bigg|^4_1 + 3\ln w-3\bigg|^4_1 \Rightarrow 3 + 3\ln -2 + c$$
Did I do this correctly?
Edit: Just realized I did it wrong. There is a asymptote at w = 3 so I now have it set up like the following:
$$\int_1^4\frac{w}{w-3}dw \rightarrow \int_1^3 \frac{w}{w-3}dw + \int_3^4\frac{w}{w-3}dw$$ $$\int_1^3 \frac{w}{w-3}dw\; \Rightarrow \;^\lim_{t\rightarrow 3^-} \int^t_1 \frac{w}{w-3}\; \Rightarrow \;^\lim_{t\rightarrow 3^-} \int^{t-3}_{-2} \frac{w}{w-3} $$ $$^\lim_{t\rightarrow 3^-} \int^{t-3}_{-2} du + ^\lim_{t\rightarrow 3^-} \int^{t-3}_{-2}\frac1udu$$ $$^\lim_{t\rightarrow 3^-}t-1+3ln(t-3)-3ln(2)$$
is this looking right?
Edit2: Through calculators I have found that $$^\lim_{t\rightarrow 3^-}t-1+3ln(t-3)-3ln(2) = -\infty$$ If this is the case, the integral is divergent rendering it impossible to integrate.
Can anyone confirm that this is correct?
Be careful with the substitutions; it's good to set $u=w-3$, but with this the integral becomes $$ \int_1^4\frac{w}{w-3}\,dw= \int_{-2}^1\frac{u+3}{u}\,du= \int_{-2}^1du+\int_{-2}^1\frac{1}{u}\,du $$ Now we see the problem more clearly: the function $f(u)=1/u$ is not defined in $0$, which is inside the interval over which we integrate. Thus we must interpret this as an improper (or generalized) integral and existence of $\int_{-2}{1}(1/u)\,du$ is equivalent to the (existence and) finiteness of both $$ \lim_{x\to0^-}\int_{-2}^x\frac{1}{u}\,du \qquad\text{and}\qquad \lim_{x\to0^+}\int_{x}^1\frac{1}{u}\,du $$ Neither is finite: for instance $$ \int_x^1\frac{1}{u}\,du=\log1-\log x\xrightarrow{x\to0^+}\infty $$
Note that an incorrect result would be obtained if one doesn't take care of the singularity: arguing like $$ \int_{-2}^1du+\int_{-2}^1\frac{1}{u}\,du=[u]_{-2}^1+[\log|u|]_{-2}^1= 1+2+\log 1-\log|{-2}|=3-\log2 $$ would be utterly wrong.