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I'm not sure how to do this, I'm also new to math.stackexchange so please excuse any novice mistakes. So anyways, here is a question I have on a summer assignment for Calculus BC (this is review from AB).

Find: $$\int {x\over(2x-1)} \, \mathrm{d}x$$

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2 Answers

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$$\int \frac{x}{2x-1} \, \mathrm{d}x = \int \frac12 \cdot \frac{2x-1+1}{2x-1} \, \mathrm{d}x= \frac{1}{2} \int 1+\frac{1}{2x-1}\, \mathrm{d}x$$

So we get $$\int \frac{x}{2x-1} \, \mathrm{d}x = \frac{1}{2} \left(x + \frac{1}{2} \ln{|2x-1|}\right) + \mathrm{c}$$

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Substitute $y = 2x-1$. Then $x = \tfrac{y+1}{2}$, and $\tfrac{dy}{dx} = 2$. Then

$$\int_{a}^b \frac{x}{2x-1} dx = \int_{\frac{a+1}{2}}^{\frac{b+1}2} \frac{\frac{y+1}{2}}y \frac{dy}{2} = \frac 14 \int_{\frac{a+1}{2}}^{\frac{b+1}2} \frac{y+1}y dy = \frac 14 \int_{\frac{a+1}{2}}^{\frac{b+1}2} 1 + \frac 1y dy.$$

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