$\int\dfrac1x\,\mathrm dx=\ln|x|+C\qquad x\ne0$ \(\int\dfrac1{ax+b}\,\mathrm dx=\dfrac1a\ln|ax+b|+

Question

$\int\dfrac1x\,\mathrm dx=\ln|x|+C\qquad x\ne0$ $\int\dfrac1{ax+b}\,\mathrm dx=\dfrac1a\ln|ax+b|+C\qquad a\ne0,\ ax+b\ne0$ The power rule excludes $n=-1$, because its denominator $n+1$ would be zero. Use the logarithm rule instead. Example Integrate $\dfrac1{3x}$ with respect to $x$, where $x\ne0$. $\int\dfrac1{3x}\,\mathrm{d}x$${}=\dfrac13\int\dfrac1x\,\mathrm{d}x$${}=\dfrac13\ln|x|+C$ Integrate $\dfrac1{2x+1}$ with respect to $x$, on an interval where $2x+1\ne0$. $\int\dfrac1{2x+1}\,\mathrm{d}x$${}=\dfrac12\ln|2x+1|+C$

Accepted Answer

$\displaystyle\int\dfrac1x\,\mathrm{d}x=\ln|x|+C$, and $\displaystyle\int\dfrac1{ax+b}\,\mathrm{d}x=\dfrac1a\ln|ax+b|+C$ for $a\ne0$

Integration of \(\frac{1}{ax+b}\)

Integration of \(\frac{1}{ax+b}\)

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