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

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Question Number 213098 by MathematicalUser2357 last updated on 30/Oct/24
∫((3x+2)/(5x^2 +2x+3))dx=?
$$\int\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x}=? \\ $$
Answered by issac last updated on 30/Oct/24
Hmmmmm..... ((3x+2)/(5x^2 +2x+3))=((3(10x+2))/(10(5x^2 +2x+3)))+(7/(5(5x^2 +2x+3))) ∫ ((3x+2)/(5x^2 +2x+3)) =∫ ((3(10x+2))/(10(5x^2 +2x+3)))+∫ (7/(5(5x^2 +2x+3))) (3/(10)) ∫ ((10x+2)/(5x^2 +2x+3))dx+(7/5) ∫ (1/(5x^2 +2x+3))dx u=5x^2 +2x+3 → du=(10x+2)dx (3/(10)) ∫ (du/u)=(3/(10))ln(u)=(3/(10))ln(5x^2 +2x+3) (A) and (7/5) ∫ (1/(5x^2 +2x+3))dx=(7/5) ∫ (1/(((√5)x+(1/( (√5))))^2 +((14)/5)))dx (√5)x+(1/( (√5)))=u → du=(√5)dx (7/(5(√5))) ∫ (5/(14(((5u^2 )/(14))+1)))du=(1/(2(√5))) ∫ (1/(((5u^2 )/(14))+1))du p=(√(5/(14)))v → dp=(√(5/(14))) dv (1/5)(√(7/2)) ∫ (1/(p^2 +1))dp=(1/5)(√(7/2)) tan^(−1) (p) =(1/5)(√(7/2)) tan^(−1) ((√(5/(14)))v)=(1/5)(√(7/2))tan^(−1) (((5x+1)/( (√(14))))) (B) ∴ Solution=A+B y(x)=(3/(10))ln(5x^2 +2x+3)+((√(14))/(10)) tan^(−1) (((5x+1)/( (√(14))))) Holy.....shit....what a terrible differantial equation XD
$$\mathrm{Hmmmmm}….. \\ $$$$\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}=\frac{\mathrm{3}\left(\mathrm{10}{x}+\mathrm{2}\right)}{\mathrm{10}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}+\frac{\mathrm{7}}{\mathrm{5}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)} \\ $$$$\int\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\:=\int\:\frac{\mathrm{3}\left(\mathrm{10}{x}+\mathrm{2}\right)}{\mathrm{10}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}+\int\:\frac{\mathrm{7}}{\mathrm{5}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)} \\ $$$$\frac{\mathrm{3}}{\mathrm{10}}\:\int\:\:\frac{\mathrm{10}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x}+\frac{\mathrm{7}}{\mathrm{5}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x} \\ $$$${u}=\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\:\:\:\rightarrow\:\mathrm{d}{u}=\left(\mathrm{10}{x}+\mathrm{2}\right)\mathrm{d}{x} \\ $$$$\frac{\mathrm{3}}{\mathrm{10}}\:\int\:\:\frac{\mathrm{d}{u}}{{u}}=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{ln}\left({u}\right)=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{ln}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\:\:\:\left(\mathrm{A}\right) \\ $$$$\mathrm{and}\:\frac{\mathrm{7}}{\mathrm{5}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x}=\frac{\mathrm{7}}{\mathrm{5}}\:\int\:\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{5}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} +\frac{\mathrm{14}}{\mathrm{5}}}\mathrm{d}{x} \\ $$$$\sqrt{\mathrm{5}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}={u}\:\:\rightarrow\:\mathrm{d}{u}=\sqrt{\mathrm{5}}\mathrm{d}{x} \\ $$$$\frac{\mathrm{7}}{\mathrm{5}\sqrt{\mathrm{5}}}\:\int\:\:\frac{\mathrm{5}}{\mathrm{14}\left(\frac{\mathrm{5}{u}^{\mathrm{2}} }{\mathrm{14}}+\mathrm{1}\right)}\mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{5}{u}^{\mathrm{2}} }{\mathrm{14}}+\mathrm{1}}\mathrm{d}{u} \\ $$$${p}=\sqrt{\frac{\mathrm{5}}{\mathrm{14}}}{v}\:\:\rightarrow\:\:\mathrm{d}{p}=\sqrt{\frac{\mathrm{5}}{\mathrm{14}}}\:\mathrm{d}{v} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\int\:\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{p}=\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left({p}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{5}}{\mathrm{14}}}{v}\right)=\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}{x}+\mathrm{1}}{\:\sqrt{\mathrm{14}}}\right)\:\:\left(\mathrm{B}\right) \\ $$$$\therefore\:\mathrm{Solution}=\mathrm{A}+\mathrm{B} \\ $$$${y}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{ln}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)+\frac{\sqrt{\mathrm{14}}}{\mathrm{10}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}{x}+\mathrm{1}}{\:\sqrt{\mathrm{14}}}\right) \\ $$$$\:\:\: \\ $$$$\mathcal{H}\mathrm{oly}…..\mathrm{shit}….\mathrm{what}\:\mathrm{a}\:\mathrm{terrible}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\:\mathrm{XD} \\ $$
Answered by MrGaster last updated on 30/Oct/24
=(1/5)∫((15x+10)/(5x^2 +2x+3))dx =(1/5)∫((15x+2+8)/(5x^2 +2x+3))dx =(1/5)(∫((15x+2)/(5x^2 +2x+3))dx+∫(8/(5x^2 +2x+3))dx) =(1/5)(∫((d(5x^2 +2x+3))/(5x^2 +2x+3))+∫(8/(5(x^2 +(2/5)x+(3/5)))dx) =(1/5)(ln∣5x^2 +2x+3∣+8∫(1/(5(x^2 +(2/5)x+(3/5))))dx) =(1/5)(ln∣5x^2 +2x+3∣+(8/5)∫(1/((x+(1/5))^2 +(((√(14))/5))^2 ))dx) =(1/5)ln∣5x^2 +2x+3∣+(8/(25))∙(5/( (√(14))))arctan(((x+(1/5))/((√(14))/5)))+C =(1/5)ln∣5x^2 +2x+3∣+(8/( (√(14))))arctan(((5x+1)/( (√(14)))))+C
$$=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{15}{x}+\mathrm{10}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{15}{x}+\mathrm{2}+\mathrm{8}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\int\frac{\mathrm{15}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx}+\int\frac{\mathrm{8}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\int\frac{{d}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}+\int\frac{\mathrm{8}}{\mathrm{5}\left({x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{5}}{x}+\frac{\mathrm{3}}{\mathrm{5}}\right.}{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{ln}\mid\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\mid+\mathrm{8}\int\frac{\mathrm{1}}{\mathrm{5}\left({x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{5}}{x}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{ln}\mid\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\mid+\frac{\mathrm{8}}{\mathrm{5}}\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{14}}}{\mathrm{5}}\right)^{\mathrm{2}} }{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\mid\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\mid+\frac{\mathrm{8}}{\mathrm{25}}\centerdot\frac{\mathrm{5}}{\:\sqrt{\mathrm{14}}}\mathrm{arctan}\left(\frac{{x}+\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\sqrt{\mathrm{14}}}{\mathrm{5}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\mid\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\mid+\frac{\mathrm{8}}{\:\sqrt{\mathrm{14}}}\mathrm{arctan}\left(\frac{\mathrm{5}{x}+\mathrm{1}}{\:\sqrt{\mathrm{14}}}\right)+{C} \\ $$
Answered by Ghisom last updated on 30/Oct/24
∫((3x+2)/(5x^2 +2x+3))dx= [t=((5x+1)/( (√(14))))] =(3/5)∫(t/(t^2 +1))dt+((√(14))/(10))∫(dt/(t^2 +1))= =(3/(10))ln (t^2 +1) +((√(14))/(10))arctan t = =(3/(10))ln (5x^2 +2x+3) +((√(14))/(10))arctan ((5x+1)/( (√(14)))) +C
$$\int\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{5}{x}+\mathrm{1}}{\:\sqrt{\mathrm{14}}}\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\frac{\sqrt{\mathrm{14}}}{\mathrm{10}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{14}}}{\mathrm{10}}\mathrm{arctan}\:{t}\:= \\ $$$$=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{ln}\:\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\:+\frac{\sqrt{\mathrm{14}}}{\mathrm{10}}\mathrm{arctan}\:\frac{\mathrm{5}{x}+\mathrm{1}}{\:\sqrt{\mathrm{14}}}\:+{C} \\ $$

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