1x, dx=ln|x|+C x 0 1ax+b, dx= 1aln|ax+b|+C a 0, ax+b 0 The power rule excludes n=-1, because its denominator n+1 would be zero. Use the logarithm rule instead. Example Integrate 13x with respect to x, where x 0. 13x,dx = 13 1x,dx = 13ln|x|+C Integrate 12x+1 with respect to x, on an interval where 2x+1 0. 12x+1,dx = 12ln|2x+1|+C

\(\int\dfrac1x\,\mathrm dx=\ln|x|+C\qquad x\ne0\)

\(\int\dfrac1{ax+b}\,\mathrm dx=\dfrac1a\ln|ax+b|+C\qquad a\ne0,\ ax+b\ne0\)

The power rule excludes \(n=-1\), because its denominator \(n+1\) would be zero. Use the logarithm rule instead.

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 \(2x+1\ne0\).

\(\int\dfrac1{2x+1}\,\mathrm{d}x\)\({}=\dfrac12\ln|2x+1|+C\)

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

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

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