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Question Number 120562 by peter frank last updated on 01/Nov/20

Answered by Dwaipayan Shikari last updated on 01/Nov/20

$$\int 1-\frac{bx+12}{x^2+bx+12}dx$$$$=x-b\int \frac{x+\frac{12}{b}}{x^2+bx+12}dx$$$$=x-\frac{b}{2}\int \frac{2x+b}{x^2+bx+12}dx+\frac{\frac{24}{b}-b}{x^2+bx+12}dx$$$$=x-\frac{b}{2}log(x^2+bx+12)-\frac{24-b^2}{2}\int \frac{1}{{(x+\frac{b}{2})}^2+{\left(\sqrt{12-\frac{b^2}{4}}\right)}^2}dx$$$$=x-\frac{b}{2}log(x^2+bx+12)-\frac{24-b^2}{2}.\frac{1}{\sqrt{12-\frac{b^2}{4}}}{tan}^{-1}\frac{2x+b}{\sqrt{48-b^2}}$$$$=x-\frac{b}{2}log(x^2+bx+12)-\frac{24-b^2}{\sqrt{48-b^2}}{tan}^{-1}\frac{2x+b}{\sqrt{48-b^2}}+C$$

Commented by peter frank last updated on 01/Nov/20

$$\mathrm{thank}\mathrm{you}$$

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