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Question Number 154186 by daus last updated on 15/Sep/21

Answered by puissant last updated on 15/Sep/21

$$\mathrm{\Omega }=\int \frac{x^3-7x^2+8x+3}{x^2-7x+12}dx$$$$=\int x-\frac{4x-3}{x^2-7x+12}dx=\frac{1}{2}{x}^2-\int \frac{4x-3}{x^2-7x+12}dx$$$$=\frac{1}{2}{x}^2-2\int \frac{2x-\frac{3}{2}-\frac{11}{2}+\frac{11}{2}}{x^2-7x+12}dx$$$$=\frac{1}{2}{x}^2-2\int \frac{2x-7}{x^2-7x+12}dx-11\int \frac{dx}{x^2-7x+12}$$$$=\frac{1}{2}{x}^2-2ln\mid x^2-7x+12\mid -11\int \frac{dx}{(x-\frac{7}{2})-\frac{49}{4}+12}$$$$=\frac{1}{2}{x}^2-2ln\mid x^2-7x+12\mid -11\int \frac{dx}{\frac{1}{4}{\left[2(x-\frac{7}{2})\right]}^2+1}$$$$u=2(x-\frac{7}{2})\to dx=\frac{du}{2}$$$$\mathrm{\Omega }=\frac{1}{2}{x}^2-2ln\mid x^2-7x+12\mid -22\int \frac{du}{1+u^2}$$$$=\frac{1}{2}{x}^2-2ln\mid x^2-7x+12\mid -22arctan\left(u\right)+C..$$$$$$$$\therefore \because \mathrm{\Omega }=\frac{1}{2}{x}^2-2ln\mid x^2-7x+12\mid -22arctan(2x-7)+C..$$

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