Evaluate the integral $\int_{1}^{4} w/(w-3) dw$ if possible

Question

Evaluate the following integral, if possible: $$\int_{1}^{4} \frac{w}{w-3} dw.$$ I am supposed to be using improper integrals so I know I should find $$ \lim_{t \rightarrow3^-} \int_{1}^{t} \frac{w}{w-3} dw + \lim_{t \rightarrow 3^+} \int_{t}^{4} \frac{w}{w-3} dw $$ because $w$ can't equal $3$. Next, I need to solve for the integral, which for some reason I can't think of how to do AT ALL. Can someone please help?

Answer 1

Hint: Notice that $\frac{w}{w-3} = \frac{w - 3 + 3}{w-3} = 1 +\frac{3}{w-3}$ and $lim_{t \to 3^{-}} ln |t - 3| = -\infty$

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Answer 2

Hint: introduce variable $u:=w-2$ to reduce the problem to that of $\int_0^t\frac1x\,dx$.