Can anyone show me how id prove this fact? I have no clue where to begin so any hints to help or full answers would be great!
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How about taking the derivative of right hand side? – Spock Jan 16 '14 at 12:48
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What's your definition of $\ln{x}$? A lot of people simply define it as the integral of $1/x$. – David H Jan 16 '14 at 13:04
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Since you suspect that $\int\frac1x\,\mathrm{d}x=\log|x|+C$, you can simply take the derivative to verify: $$ \frac{\mathrm{d}}{\mathrm{d}x}\log(x)=\frac1x\quad\text{when }x\gt0 $$ $$ \frac{\mathrm{d}}{\mathrm{d}x}\log(-x)=\frac1x\quad\text{when }x\lt0 $$ If you don't suspect $\log|x|+C$, you can use $$ \int x^n\,\mathrm{d}x=\frac1{n+1}\left(x^{n+1}-1\right)+C $$ as in this answer.
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Perhaps $$ \int \frac{1}{x}\;dx = \ln|x|+C $$ is misleading. First, it is false for complex $x$. Second, for real $x$, it is shorthand for $$ \int\frac{1}{x}\;dx = \ln x + C \qquad\text{on intervals where $x>0$ and} \\ \int\frac{1}{x}\;dx = \ln(-x) + D \qquad\text{on intervals where $x<0$.} $$ The constants $C$ and $D$ need not be the same.
GEdgar
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