Splitting the Numerator

Splitting the Numerator

Split-up method

\[\mathrm{f}(x)=A\mathrm{g}'(x)+B\]

\[\int\frac{\mathrm{f}(x)}{\mathrm{g}(x)}\,\mathrm{d}x=A\int\frac{\mathrm{g}'(x)}{\mathrm{g}(x)}\,\mathrm{d}x+B\int\frac{1}{\mathrm{g}(x)}\,\mathrm{d}x\]

\[\int\frac{\mathrm{f}(x)}{\sqrt{\mathrm{g}(x)}}\,\mathrm{d}x=A\int\frac{\mathrm{g}'(x)}{\sqrt{\mathrm{g}(x)}}\,\mathrm{d}x+B\int\frac{1}{\sqrt{\mathrm{g}(x)}}\,\mathrm{d}x\]

IntegrandRecognisable termRemaining standard term
\(\dfrac{\mathrm{f}(x)}{\mathrm{g}(x)}\)\(A\dfrac{\mathrm{g}'(x)}{\mathrm{g}(x)}\)\(\dfrac{B}{\mathrm{g}(x)}\)
\(\dfrac{\mathrm{f}(x)}{\sqrt{\mathrm{g}(x)}}\)\(A\dfrac{\mathrm{g}'(x)}{\sqrt{\mathrm{g}(x)}}\)\(\dfrac{B}{\sqrt{\mathrm{g}(x)}}\)

Worked pattern

\(8x+3=(8x+2)+1\)

\(\int\frac{8x+3}{4x^2+2x+1}\,\mathrm dx=\int\frac{8x+2}{4x^2+2x+1}\,\mathrm dx+\int\frac1{4x^2+2x+1}\,\mathrm dx\)

\(u=4x^2+2x+1\Rightarrow\mathrm du=(8x+2)\,\mathrm dx\)

\(4x^2+2x+1=4\left[\left(x+\frac14\right)^2+\left(\frac{\sqrt3}4\right)^2\right]\)

\(=\ln|4x^2+2x+1|+\frac1{\sqrt3}\tan^{-1}\left(\frac{4x+1}{\sqrt3}\right)+C\)

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Answer:\(\ln|4x^2+2x+1|+\frac1{\sqrt3}\tan^{-1}\left(\frac{4x+1}{\sqrt3}\right)+C\)

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