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integrate-with-respect-to-x-dx-9x-2-6x-10-




Question Number 21077 by j.masanja06@gmail.com last updated on 11/Sep/17
integrate with respect to x  ∫(dx/(9x^2  +6x+10))
$${integrate}\:{with}\:{respect}\:{to}\:{x} \\ $$$$\int\frac{{dx}}{\mathrm{9}{x}^{\mathrm{2}} \:+\mathrm{6}{x}+\mathrm{10}} \\ $$
Answered by Joel577 last updated on 12/Sep/17
9x^2  + 6x + 1 + 9  I = ∫ (dx/((3x + 1)^2  + 9))  Let u = 3x +1  →  du = 3 dx  I = (1/3)∫ (du/(u^2  + 9))     = (1/9)tan^(−1) ((u/3)) + C     = (1/9)tan^(−1) (x + (1/3)) + C
$$\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{1}\:+\:\mathrm{9} \\ $$$${I}\:=\:\int\:\frac{{dx}}{\left(\mathrm{3}{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{9}} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{3}{x}\:+\mathrm{1}\:\:\rightarrow\:\:{du}\:=\:\mathrm{3}\:{dx} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{du}}{{u}^{\mathrm{2}} \:+\:\mathrm{9}} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{9}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{3}}\right)\:+\:{C} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{9}}\mathrm{tan}^{−\mathrm{1}} \left({x}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:{C} \\ $$

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