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Question Number 124540 by micelle last updated on 04/Dec/20

$$\int \frac{4x+9}{x^2+6x+10}dx$$

Answered by Ar Brandon last updated on 04/Dec/20

$$\mathcal{I}=\int \frac{4\mathrm{x}+9}{{\mathrm{x}}^2+6\mathrm{x}+10}\mathrm{dx}$$$$4\mathrm{x}+9=\lambda \left\{\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2+6\mathrm{x}+10)\right\}+\mu =\lambda (2\mathrm{x}+6)+\mu$$$$\mathrm{comparing}\mathrm{coefs},\lambda =2,\mu =-3$$$$4\mathrm{x}+9=2(2\mathrm{x}+6)-3$$$$\mathcal{I}=\int \{\frac{2(2\mathrm{x}+6)}{{\mathrm{x}}^2+6\mathrm{x}+10}-\frac{3}{{\mathrm{x}}^2+6\mathrm{x}+10}\}\mathrm{dx}$$$$=2\ln ({\mathrm{x}}^2+6\mathrm{x}+10)-{3\tan }^{-1}(\mathrm{x}+3)+\mathcal{C}$$

Answered by mathmax by abdo last updated on 04/Dec/20

$$\int \frac{4\mathrm{x}+9}{{\mathrm{x}}^2+6\mathrm{x}+10}\mathrm{dx}=\int \frac{2\mathrm{x}+6+2\mathrm{x}+3}{{\mathrm{x}}^2+6\mathrm{x}+10}\mathrm{dx}$$$$=\int \frac{2\mathrm{x}+6}{{\mathrm{x}}^2+6\mathrm{x}+10}\mathrm{dx}+\int \frac{2\mathrm{x}+6-3}{{\mathrm{x}}^2+6\mathrm{x}+10}\mathrm{dx}$$$$=2\ln ({\mathrm{x}}^2+6\mathrm{x}+10)-3\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+6\mathrm{x}+9+1}$$$$=2\ln ({\mathrm{x}}^2+6\mathrm{x}+10)-3\int \frac{\mathrm{dx}}{{(\mathrm{x}+3)}^2+1}(\to \mathrm{x}+3=\mathrm{u})$$$$=2\ln ({\mathrm{x}}^2+6\mathrm{x}+10)-3\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}$$$$=2\ln ({\mathrm{x}}^2+6\mathrm{x}+10)-3\arctan (\mathrm{x}+3)+\mathrm{C}$$

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