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Question Number 124261 by Ar Brandon last updated on 02/Dec/20

∫((2x^2 +5x+9)/((x+1)(√(x^2 +x+1))))dx

$$\int\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{9}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\mathrm{dx} \\ $$

Commented by MJS_new last updated on 02/Dec/20

I think something went wrong...

$$\mathrm{I}\:\mathrm{think}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}... \\ $$

Commented by Ar Brandon last updated on 02/Dec/20

I get (√(x^2 +x+1))+(1/( (√3)))tan^(−1) (((2x+1)/( (√3)))) −6ln∣((1−x+2(√(x^2 +x+1)))/(2(x+1)))∣+K Any approval, please ?

$$\mathrm{I}\:\mathrm{get} \\ $$$$\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$−\mathrm{6ln}\mid\frac{\mathrm{1}−\mathrm{x}+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\mid+\mathrm{K} \\ $$$${Any}\:{approval},\:{please}\:? \\ $$

Answered by MJS_new last updated on 02/Dec/20

∫((2x^2 +5x+9)/((x+1)(√(x^2 +x+1))))dx= [t=x+(1/2) → dx=dt] =4∫((2t^2 +3t+7)/((2t+1)(√(4t^2 +3))))dt= [u=(((√3)(2t+(√(4t^2 +3))))/3) → dt=((√(3(4t^2 +3)))/(2(2t+(√(4t^2 +3)))))] =∫((3(√3)u^4 +18u^3 +50(√3)u^2 −18u+3(√3))/(2u^2 (3u^2 +2(√3)u−3)))du= =∫(((√3)/2)−(6/(u+(√3)))+(2/u)+(6/(u−((√3)/3)))−((√3)/(2u^2 )))du= =((√3)/2)u−6ln (u+(√3))+2ln u +6ln (u−((√3)/3)) +((√3)/(2u))= =(((√3)(u^2 +1))/(2u))+6ln ((3u−(√3))/(u+(√3))) +2ln u = ... =2(√(x^2 +x+1))+6ln ((x−1+2(√(x^2 +x+1)))/(x+1)) +2ln (2x+1+2(√(x^2 +x+1))) +C

$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{9}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\mathrm{4}\int\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{7}}{\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}\right)}{\mathrm{3}}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{3}\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}\right)}}{\mathrm{2}\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}\right)}\right] \\ $$$$=\int\frac{\mathrm{3}\sqrt{\mathrm{3}}{u}^{\mathrm{4}} +\mathrm{18}{u}^{\mathrm{3}} +\mathrm{50}\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{18}{u}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}{u}^{\mathrm{2}} \left(\mathrm{3}{u}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{u}−\mathrm{3}\right)}{du}= \\ $$$$=\int\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{6}}{{u}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{{u}}+\frac{\mathrm{6}}{{u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{u}^{\mathrm{2}} }\right){du}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}−\mathrm{6ln}\:\left({u}+\sqrt{\mathrm{3}}\right)+\mathrm{2ln}\:{u}\:+\mathrm{6ln}\:\left({u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{u}}= \\ $$$$=\frac{\sqrt{\mathrm{3}}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{u}}+\mathrm{6ln}\:\frac{\mathrm{3}{u}−\sqrt{\mathrm{3}}}{{u}+\sqrt{\mathrm{3}}}\:+\mathrm{2ln}\:{u}\:= \\ $$$$... \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\mathrm{6ln}\:\frac{{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{{x}+\mathrm{1}}\:+\mathrm{2ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)\:+{C} \\ $$

Commented by Ar Brandon last updated on 02/Dec/20

Thanks Sir, I found my mistake.

Commented by Ar Brandon last updated on 02/Dec/20

��Lol, OK ��

Commented by Ar Brandon last updated on 02/Dec/20

By the way. I left a message for you privately. Can you please check ?��

Commented by Ar Brandon last updated on 02/Dec/20

I don't think so, bro. Perhaps you're talking about the difference in sign. It's due to the absolute value of "u" I think.

Commented by Ar Brandon last updated on 02/Dec/20

OK. Thanks for your opinion, Sir. I'll check and post my solution.

Commented by Ar Brandon last updated on 02/Dec/20

Sir, I don't understand how you decide on your change of variables everytime. And they sound pretty fast��

Commented by MJS_new last updated on 02/Dec/20

I hate to deal with sinh and cosh so I usually use these: (√(ax^2 ±b)) ⇒ t=(((√a)x+(√(ax^2 ±b)))/( (√b))) ⇔ x=(((√b)(t^2 ∓1))/(2(√a)t)) → dx=((√(b(ax^2 ±b)))/(ax+(√(a(ax^2 ±b)))))

$$\mathrm{I}\:\mathrm{hate}\:\mathrm{to}\:\mathrm{deal}\:\mathrm{with}\:\mathrm{sinh}\:\mathrm{and}\:\mathrm{cosh}\:\mathrm{so}\:\mathrm{I}\:\mathrm{usually} \\ $$$$\mathrm{use}\:\mathrm{these}: \\ $$$$\sqrt{{ax}^{\mathrm{2}} \pm{b}}\:\Rightarrow\:{t}=\frac{\sqrt{{a}}{x}+\sqrt{{ax}^{\mathrm{2}} \pm{b}}}{\:\sqrt{{b}}}\:\Leftrightarrow\:{x}=\frac{\sqrt{{b}}\left({t}^{\mathrm{2}} \mp\mathrm{1}\right)}{\mathrm{2}\sqrt{{a}}{t}}\:\rightarrow\:{dx}=\frac{\sqrt{{b}\left({ax}^{\mathrm{2}} \pm{b}\right)}}{{ax}+\sqrt{{a}\left({ax}^{\mathrm{2}} \pm{b}\right)}} \\ $$

Commented by Ar Brandon last updated on 02/Dec/20

Oh, I see ! ��

Commented by Dwaipayan Shikari last updated on 02/Dec/20

One or two more sir :)

$$\left.{One}\:{or}\:{two}\:{more}\:{sir}\::\right) \\ $$

Commented by Dwaipayan Shikari last updated on 02/Dec/20

I have told that sir may share one or more examples regarding this process ��

Commented by Ar Brandon last updated on 02/Dec/20

No, I meant on telegram. Sorry I didn't precise.

Answered by Ar Brandon last updated on 02/Dec/20

I=∫((2x^2 +5x+9)/((x+1)(√(x^2 +x+1))))dx ((2x^2 +5x+9)/(x+1))=((2(x+1)^2 +(x+1)+6)/(x+1))=2(x+1)+1+(6/(x+1)) I=∫{2(x+1)+1+(6/(x+1))}∙(dx/( (√(x^2 +x+1)))) =2∫((x+1)/( (√(x^2 +x+1))))dx+∫(dx/( (√(x^2 +x+1))))+6∫(dx/((x+1)(√(x^2 +x+1)))) =∫((2x+1)/( (√(x^2 +x+1))))dx+2∫(dx/( (√(x^2 +x+1))))+6∫(dx/((x+1)(√(x^2 +x+1)))) =2(√(x^2 +x+1))+2ln∣(x+(1/2))+(√(x^2 +x+1))∣+6J For J, Let (1/(x+1))=u ⇒ −(dx/((x+1)^2 ))=du ⇒ dx=−(du/u^2 ) ⇒ x=(1/u)−1, x^2 +x+1=(1/u^2 )−(1/u)+1 J=−∫(u/( (√((1/u^2 )−(1/u)+1))))∙(du/u^2 )=−∫(du/( (√(1−u+u^2 )))) =−∫(du/( (√((u−(1/2))^2 +(((√3)/2))^2 ))))=−sinh^(−1) (((u−1/2)/( (√3)/2)))+C =−ln∣(u−(1/2))+(√((u−(1/2))^2 +(((√3)/2))^2 ))∣+C =−ln∣((1/(x+1))−(1/2))+(√((1/((x+1)^2 ))−(1/(x+1))+1))∣+C =−ln∣((1−x+2(√(x^2 +x+1)))/(2(x+1)))∣+C ⇒I=2(√(x^2 +x+1))+2ln∣(x+(1/2))+(√(x^2 +x+1))∣−6ln∣((1−x+2(√(x^2 +x+1)))/(2(x+1)))∣+K

$$\mathcal{I}=\int\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{9}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\mathrm{dx}\: \\ $$$$\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{9}}{\mathrm{x}+\mathrm{1}}=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{6}}{\mathrm{x}+\mathrm{1}}=\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}+\mathrm{1}} \\ $$$$\mathcal{I}=\int\left\{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}+\mathrm{1}}\right\}\centerdot\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}} \\ $$$$\:\:\:=\mathrm{2}\int\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\mathrm{dx}+\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}+\mathrm{6}\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}} \\ $$$$\:\:\:=\int\frac{\mathrm{2x}+\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\mathrm{dx}+\mathrm{2}\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}+\mathrm{6}\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}} \\ $$$$\:\:\:=\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\mathrm{2ln}\mid\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mid+\mathrm{6}\mathcal{J} \\ $$$$\mathrm{For}\:\mathcal{J}, \\ $$$$\mathrm{Let}\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}=\mathrm{u}\:\Rightarrow\:−\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{du}\:\Rightarrow\:\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{u}}−\mathrm{1},\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{u}}+\mathrm{1} \\ $$$$\mathcal{J}=−\int\frac{\mathrm{u}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{u}}+\mathrm{1}}}\centerdot\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} }=−\int\frac{\mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{u}+\mathrm{u}^{\mathrm{2}} }} \\ $$$$\:\:\:\:=−\int\frac{\mathrm{du}}{\:\sqrt{\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }}=−\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{u}−\mathrm{1}/\mathrm{2}}{\:\sqrt{\mathrm{3}}/\mathrm{2}}\right)+\mathrm{C} \\ $$$$\:\:\:\:=−\mathrm{ln}\mid\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\mid+\mathrm{C} \\ $$$$\:\:\:\:=−\mathrm{ln}\mid\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}+\mathrm{1}}\mid+\mathrm{C} \\ $$$$\:\:\:\:=−\mathrm{ln}\mid\frac{\mathrm{1}−\mathrm{x}+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\mid+\mathrm{C} \\ $$$$\Rightarrow\mathcal{I}=\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\mathrm{2ln}\mid\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mid−\mathrm{6ln}\mid\frac{\mathrm{1}−\mathrm{x}+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\mid+\mathrm{K} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Dec/20

=∫(((2x+1))/( (√(x^2 +x+1))))dx +∫((2x+8)/((x+1)(√(x^2 +x+1))))dx =2(√(x^2 +x+1)) +2∫(1/( (√(x^2 +x+1))))+6∫(1/( (x+1)(√(x^2 +x+1))))dx (x+1)=t =2(√(x^2 +x+1)) +2∫(1/( (√((x+(1/2))^2 +(((√3)/2))^2 ))))+6∫(1/(t(√(t^2 −t+1))))dt, t=(1/u) =2(√(x^2 +x+1)) +2log(x+(1/2)+(√(x^2 +x+1)))−6∫(1/( (√(u^2 −u+1))))du =2((√(x^2 +x+1))+log(x+(1/2)+(√(x^2 +x+1))))−6log(u−(1/2)+(√(u^2 −u+1))) 2(√(x^2 +x+1))+log(x+(1/2)+(√(x^2 +x+1)))−6log((1/(x+1))−(1/2)+(√((1/((x+1)^2 ))−(1/((x+1)))−1)))+C

$$=\int\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}\:+\int\frac{\mathrm{2}{x}+\mathrm{8}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+\mathrm{6}\int\frac{\mathrm{1}}{\:\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}\:\:\:\:\:\:\:\left({x}+\mathrm{1}\right)={t} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }}+\mathrm{6}\int\frac{\mathrm{1}}{{t}\sqrt{{t}^{\mathrm{2}} −{t}+\mathrm{1}}}{dt},\:\:\:{t}=\frac{\mathrm{1}}{{u}} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{2}{log}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)−\mathrm{6}\int\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} −{u}+\mathrm{1}}}{du} \\ $$$$=\mathrm{2}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+{log}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)\right)−\mathrm{6}{log}\left({u}−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right) \\ $$$$\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+{log}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)−\mathrm{6}{log}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}−\mathrm{1}}\right)+{C} \\ $$

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