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8-2x-2-3x-5-dx-




Question Number 123314 by aurpeyz last updated on 24/Nov/20
∫(8/(2x^2 +3x+5))dx
$$\int\frac{\mathrm{8}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{5}}{dx} \\ $$
Answered by mathmax by abdo last updated on 24/Nov/20
I =8 ∫  (dx/(2(x^2  +(3/2)x+(5/2))))=4∫   (dx/(x^2  +2.(3/4)x+(9/(16))+(5/2)−(9/(16))))  =4∫   (dx/((x+(3/4))^2 +((31)/(16)))) =_(x+(3/4)=((√(31))/4)u)    4×((16)/(31))∫  (1/(u^2  +1)).((√(31))/4)du  =((16)/( (√(31)))) arctan(u) +C =((16)/( (√(31))))arctan(((4x+3)/( (√(31))))) +C
$$\mathrm{I}\:=\mathrm{8}\:\int\:\:\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{5}}{\mathrm{2}}\right)}=\mathrm{4}\int\:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}.\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}+\frac{\mathrm{9}}{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$$=\mathrm{4}\int\:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{31}}{\mathrm{16}}}\:=_{\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}=\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}\mathrm{u}} \:\:\:\mathrm{4}×\frac{\mathrm{16}}{\mathrm{31}}\int\:\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}.\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}\mathrm{du} \\ $$$$=\frac{\mathrm{16}}{\:\sqrt{\mathrm{31}}}\:\mathrm{arctan}\left(\mathrm{u}\right)\:+\mathrm{C}\:=\frac{\mathrm{16}}{\:\sqrt{\mathrm{31}}}\mathrm{arctan}\left(\frac{\mathrm{4x}+\mathrm{3}}{\:\sqrt{\mathrm{31}}}\right)\:+\mathrm{C} \\ $$
Commented by aurpeyz last updated on 25/Nov/20
thank you
$${thank}\:{you} \\ $$

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