201907260633 Exercise 3.1.1

Evaluate the indefinite integral:

$\displaystyle{\int\frac{x}{1-2x}}\,\mathrm{d}x$

Solution.

(bad example)

By change of variable, set $u=1-2x$ ,

then $x=\displaystyle{\frac{1-u}{2}}$ .

$\begin{aligned} \displaystyle{\frac{\mathrm{d}u}{\mathrm{d}x}} & =\displaystyle{\frac{\mathrm{d}(1-2x)}{\mathrm{d}x}}=-2 \\ \mathrm{d}x & = -\frac{1}{2}\,\mathrm{d}u \end{aligned}$ .

$\begin{aligned} & \quad \int \frac{x}{1-2x}\,\mathrm{d}x\\ & = \int \frac{\frac{1-u}{2}}{u} \bigg( -\frac{1}{2} \,\mathrm{d}u \bigg) \\ & = -\frac{1}{4} \int \bigg( \frac{1}{u} -1 \bigg) \,\mathrm{d}u \\ & = -\frac{1}{4} \big( \ln |u| - u + C \big) \\ & = -\frac{1}{4}\ln |1-2x| +\frac{1}{4} (1-2x) -\frac{1}{4}C \\ & = -\frac{1}{4}\ln |1-2x| -\frac{x}{2} + \frac{1}{4}(1-C) \\ & = -\frac{1}{4}\ln |1-2x| -\frac{x}{2} + C' \end{aligned}$

(better example)

Express $x$ in terms of $1-2x\enspace (=u)$ ,

$\begin{aligned} 1-2x & = u \\ x & = \frac{1}{2} (1-u) = \frac{1}{2} [1-(1-2x)] \end{aligned}$

Then

$\begin{aligned} & \quad \int \frac{x}{1-2x} \,\mathrm{d}x\\ & = \frac{1}{2} \int \frac{1-(1-2x)}{1-2x} \,\mathrm{d}x \\ & = \frac{1}{2} \Bigg\{ \int \frac{\mathrm{d}x}{1-2x} - \int \mathrm{d}x \Bigg\} \\ & = \frac{1}{2} \Bigg\{ \bigg(\frac{\ln |1-2x|}{-2} +C' \bigg) - \Big( x+C'' \Big) \Bigg\} \\ & = -\frac{1}{4}\ln |1-2x| - \frac{x}{2} + C \end{aligned}$

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