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Question

# $\int \frac{a{x}^{2}+bx+c}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}dx,\mathrm{where}a,b,c\mathrm{are}\mathrm{distinct}$

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}I=\int \frac{a{x}^{2}+bx+c}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}dx\phantom{\rule{0ex}{0ex}}\mathrm{Let}\frac{a{x}^{2}+bx+c}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}=\frac{A}{x\mathit{-}a}+\frac{B}{x\mathit{-}b}+\frac{C}{x-c}\phantom{\rule{0ex}{0ex}}⇒a{x}^{\mathit{2}}+bx+c=A\left(x\mathit{-}b\right)\left(x\mathit{-}c\right)+B\left(x\mathit{-}c\right)\left(x\mathit{-}a\right)+C\left(x\mathit{-}a\right)\left(x\mathit{-}b\right)\phantom{\rule{0ex}{0ex}}⇒a{x}^{\mathit{2}}+bx+c=A\left[{x}^{\mathit{2}}\mathit{-}\left(b+c\right)x+bc\right]+B\left[{x}^{\mathit{2}}\mathit{-}\left(c+a\right)x+ca\right]+C\left[{x}^{\mathit{2}}\mathit{-}\left(a+b\right)x+ab\right]\phantom{\rule{0ex}{0ex}}⇒a{x}^{\mathit{2}}+bx+c=\left(A+B+C\right){x}^{\mathit{2}}\mathit{-}\left[A\left(b+c\right)+B\left(c+a\right)+C\left(a+b\right)\right]x+Abc+Bca+Cab\phantom{\rule{0ex}{0ex}}\mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a=A+B+C.....\left(1\right)\phantom{\rule{0ex}{0ex}}b=-\left[A\left(b+c\right)+B\left(c+a\right)+C\left(a+b\right)\right].....\left(2\right)\phantom{\rule{0ex}{0ex}}c=Abc+\mathrm{B}ca+\mathrm{C}ab\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}A=\frac{{a}^{3}+ab+c}{\left(a-b\right)\left(a-c\right)}\phantom{\rule{0ex}{0ex}}B=\frac{a{b}^{2}+{b}^{2}+c}{\left(b-a\right)\left(b-c\right)}\phantom{\rule{0ex}{0ex}}C=\frac{a{c}^{2}+bc+c}{\left(c\mathit{-}a\right)\left(c\mathit{-}b\right)}\phantom{\rule{0ex}{0ex}}\therefore I=\int \left[\frac{{a}^{\mathit{3}}\mathit{+}ab\mathit{+}c}{\left(a\mathit{-}b\right)\left(a\mathit{-}c\right)}×\frac{1}{x\mathit{-}a}+\frac{a{b}^{2}+{b}^{2}+c}{\left(b-a\right)\left(b-c\right)}×\frac{1}{x\mathit{-}b}+\frac{a{c}^{2}+bc+c}{\left(c\mathit{-}a\right)\left(c\mathit{-}b\right)}×\frac{1}{x-c}\right]dx\phantom{\rule{0ex}{0ex}}=\frac{{a}^{\mathit{3}}+ab+c}{\left(a\mathit{-}b\right)\left(a\mathit{-}c\right)}\mathrm{log}\left|x-a\right|+\frac{a{b}^{2}+{b}^{2}+c}{\left(b-a\right)\left(b-c\right)}\mathrm{log}\left|x-b\right|+\frac{a{c}^{2}+bc+c}{\left(c\mathit{-}a\right)\left(c\mathit{-}b\right)}\mathrm{log}\left|x-c\right|+K\phantom{\rule{0ex}{0ex}}$

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