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Question Number 187703 by Tawa11 last updated on 20/Feb/23

∫ (1/(5x^2 − 2x − 4)) dx

$$\int\:\frac{\mathrm{1}}{\mathrm{5x}^{\mathrm{2}} \:\:−\:\:\mathrm{2x}\:\:−\:\:\mathrm{4}}\:\mathrm{dx} \\ $$

Answered by MikeH last updated on 20/Feb/23

= (1/5)∫(1/(x^2 −(2/5)x−(4/5))) dx = (1/5)∫(1/((x−(1/5))^2 −(1/(25))−(4/5)))dx = (1/5)∫(1/((x−(1/5))^2 −((21)/(25)))) dx = (1/5)∫(1/((x−(1/5))^2 −(((√(21))/5))^2 ))dx = (1/5)∫(1/(u^2 −a^2 ))du with u = (x−(1/5)) I = (1/5)∫(1/(u^2 −a^2 ))du = (1/(10a))ln(((u−a)/(u+a)))+c ⇒ I =(1/(2(√(21)))) ln(((x−((1+(√(21)))/5))/(x+((1−(√(21)))/5))))+c I =(1/(2(√(21))))ln(((5x−(1+(√(21))))/(5x+(1−(√(21))))))+c please correct me if I′m wrong.

$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{5}}{x}−\frac{\mathrm{4}}{\mathrm{5}}}\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{\left({x}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{25}}−\frac{\mathrm{4}}{\mathrm{5}}}{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{\left({x}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} −\frac{\mathrm{21}}{\mathrm{25}}}\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{\left({x}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{21}}}{\mathrm{5}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} −{a}^{\mathrm{2}} }{du}\: \\ $$$$\mathrm{with}\:{u}\:=\:\left({x}−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} −{a}^{\mathrm{2}} }{du}\:=\:\frac{\mathrm{1}}{\mathrm{10}{a}}\mathrm{ln}\left(\frac{{u}−{a}}{{u}+{a}}\right)+{c} \\ $$$$\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{21}}}\:\mathrm{ln}\left(\frac{{x}−\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{5}}}{{x}+\frac{\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{5}}}\right)+{c} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{21}}}\mathrm{ln}\left(\frac{\mathrm{5}{x}−\left(\mathrm{1}+\sqrt{\mathrm{21}}\right)}{\mathrm{5}{x}+\left(\mathrm{1}−\sqrt{\mathrm{21}}\right)}\right)+{c} \\ $$$$\mathrm{please}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{I}'\mathrm{m}\:\mathrm{wrong}.\: \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 20/Feb/23

⇒I=(1/(2(√(21))))ln∣((x−((1+(√(21)))/5))/(x−((1−(√(21)))/5)))∣+c

$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{21}}}\mathrm{ln}\mid\frac{{x}−\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{5}}}{{x}−\frac{\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{5}}}\mid+{c} \\ $$

Answered by Nimnim111118 last updated on 21/Feb/23

I=∫((5dx)/(25x^2 −10x−20))=∫((5dx)/((5x−1)^2 −((√(21)))^2 )) =(1/( 2(√(21))))ln(((5x−1−(√(21)))/(5x−1+(√(21)))))+C

$${I}=\int\frac{\mathrm{5}{dx}}{\mathrm{25}{x}^{\mathrm{2}} −\mathrm{10}{x}−\mathrm{20}}=\int\frac{\mathrm{5}{dx}}{\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{21}}\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{21}}}{ln}\left(\frac{\mathrm{5}{x}−\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{5}{x}−\mathrm{1}+\sqrt{\mathrm{21}}}\right)+{C} \\ $$

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