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Question Number 125440 by bramlexs22 last updated on 11/Dec/20

$$\int \frac{2x^2-3x-3}{(x-1)(x^2-2x+5)}dx$$

Answered by Ar Brandon last updated on 11/Dec/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{2\mathrm{x}}^2-3\mathrm{x}-3}{(\mathrm{x}-1)({\mathrm{x}}^2-2\mathrm{x}+5)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{bx}+\mathrm{c}}{{\mathrm{x}}^2-2\mathrm{x}+5}$$$$=\frac{\mathrm{a}({\mathrm{x}}^2-2\mathrm{x}+5)+(\mathrm{bx}+\mathrm{c})(\mathrm{x}-1)}{(\mathrm{x}-1)({\mathrm{x}}^2-2\mathrm{x}+5)}$$$$\mathrm{set}\mathrm{x}-1=0\Rightarrow -4=4\mathrm{a}\Rightarrow \mathrm{a}=-1$$$$\mathrm{Comparing}\mathrm{coefficients}\mathrm{of}{\mathrm{x}}^2\mathrm{we}\mathrm{get};$$$$\mathrm{a}+\mathrm{b}=2\Rightarrow \mathrm{b}=3$$$$\mathrm{Comparing}\mathrm{constant}\mathrm{terms}\mathrm{we}\mathrm{get};$$$$5\mathrm{a}-\mathrm{c}=-3\Rightarrow \mathrm{c}=-2$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{-1}{\mathrm{x}-1}+\frac{3\mathrm{x}-2}{{\mathrm{x}}^2-2\mathrm{x}+5}$$$$3\mathrm{x}-2=\lambda \left\{\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2-2\mathrm{x}+5)\right\}+\mu =\lambda (2\mathrm{x}-2)+\mu$$$$\mathrm{Comparing}\mathrm{coefs}2\lambda =3\Rightarrow \lambda =\frac{3}{2},\mu -2\lambda =-2\Rightarrow \mu =1$$$$3\mathrm{x}-2=\frac{3}{2}(2\mathrm{x}-2)+1$$$$\Rightarrow \int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int \{\frac{-1}{\mathrm{x}-1}+\frac{3}{2}\cdot \frac{2\mathrm{x}-2}{{\mathrm{x}}^2-2\mathrm{x}+5}+\frac{1}{{\mathrm{x}}^2-2\mathrm{x}+5}\}\mathrm{dx}$$

Commented by Dwaipayan Shikari last updated on 11/Dec/20

$$\int -\frac{1}{x-1}+\frac{3}{2}\int \frac{2x-2}{x^2-2x+5}+\int \frac{1}{x^2-2x+5}dx$$$$=\frac{3}{2}log(x^2-2x+5)-log(x-1)+\frac{1}{2}{tan}^{-1}\frac{x-1}{2}+C$$

Commented by Ar Brandon last updated on 11/Dec/20

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Answered by john_santu last updated on 11/Dec/20

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