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Question Number 31787 by neel1974 last updated on 14/Mar/18

$$\int \frac{4x-3}{x^2+3x+8}dx$$

Answered by sma3l2996 last updated on 14/Mar/18

$$A=\int \frac{4x-3}{x^2+3x+8}dx=2\int \frac{2x-3/2}{x^2+3x+8}dx$$$$=2\int (\frac{2x+3}{x^2+3x+8}-\frac{3+3/2}{x^2+3x+8})dx$$$$=2ln(x^2+3x+8)-9\int \frac{dx}{x^2+3x+8}+c$$$$\int \frac{dx}{x^2+3x+8}=\int \frac{dx}{x^2+2\times \frac{3}{2}\times x+{\left(\frac{3}{2}\right)}^2-\frac{9}{4}+8}$$$$=\int \frac{dx}{{(x+\frac{3}{2})}^2+\frac{23}{4}}=\frac{4}{23}\int \frac{dx}{{\left(\frac{2x+3}{\sqrt{23}}\right)}^2+1}$$$$lett=\frac{2x+3}{\sqrt{23}}\Rightarrow dt=\frac{2}{\sqrt{23}}dx$$$$\int \frac{dx}{x^2+3x+8}=\frac{2\sqrt{23}}{23}\int \frac{dt}{t^2+1}=\frac{2\sqrt{23}}{23}{tan}^{-1}\left(\frac{2x+3}{\sqrt{23}}\right)+k$$$$so$$$$A=2ln(x^2+3x+8)-\frac{18\sqrt{23}}{23}{tan}^{-1}\left(\frac{2x+3}{\sqrt{23}}\right)+C$$

Answered by ajfour last updated on 14/Mar/18

$$I=2\int \frac{2x+3}{x^2+3x+8}dx-9\int \frac{dx}{{(x+\frac{3}{2})}^2+{\left(\frac{\sqrt{23}}{2}\right)}^2}$$$$I=2\ln \mid x^2+3x+8\mid -\frac{18}{\sqrt{23}}{\tan }^{-1}\left(\frac{2x+3}{\sqrt{23}}\right)+c.$$

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