Evaluate-x-2x-2-4x-7-dx-

Question Number 179511 by Acem last updated on 30/Oct/22

$${Evaluate}\:\int\frac{{x}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}}}\:{dx} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 30/Oct/22

=(1/4)∫((4x−4)/( (√(2x^2 −4x−7))))dx+∫(dx/( (√(2(x−1)^2 −9)))) =((√(2x^2 −4x−7))/2)+((√2)/2)arcsin((((√2)(x−1))/( 3)))+C

$$=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{4}{x}−\mathrm{4}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}}}{dx}+\int\frac{{dx}}{\:\sqrt{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}}} \\ $$$$=\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arcsin}\left(\frac{\sqrt{\mathrm{2}}\left({x}−\mathrm{1}\right)}{\:\mathrm{3}}\right)+{C} \\ $$

Commented by Acem last updated on 30/Oct/22

$${Thanks}! \\ $$

Commented by Ar Brandon last updated on 30/Oct/22

You're welcome!

Commented by Frix last updated on 31/Oct/22

∫(dx/( (√(2(x−1)^2 −9))))=((√2)/2)cosh^(−1) (((√2)(x−1))/3)

$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cosh}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{2}}\left({x}−\mathrm{1}\right)}{\mathrm{3}} \\ $$

Commented by Ar Brandon last updated on 08/Nov/22

You're right. Thanks

Answered by Acem last updated on 30/Oct/22

∫(x/( (√(2x^2 −4x−7)))) dx 2x^2 −4x−7= 2(x^2 −2x+1−(9/2))= 2(x−1)^2 −9 x−1= (3/( (√2))) sec θ , dx= (3/( (√2))) sec θ tan θ dθ (√(2x^2 −4x−7 ))= 3 (√(sec^2 θ−1 ))= 3 ∣tan θ∣= 3 tan θ I= (1/( (√2))) ∫ (1+(3/( (√2))) sec θ)sec θ dθ I= (1/( (√2))) ln ∣sec θ+tan θ∣ + (3/2) tan θ + c Since sec θ= (((√2) (x−1))/3) ⇒ tan θ= ((√(2 (x−1)^2 −9))/3) I= (1/( (√2))) ln ∣(((√2) (x−1) + (√(2x^2 −4x−7 )) )/3)∣+(1/2) (√(2x^2 −4x−7 )) + c

$$\int\frac{{x}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}}}\:{dx} \\ $$$$\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}=\:\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−\frac{\mathrm{9}}{\mathrm{2}}\right)=\:\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$$\:{x}−\mathrm{1}=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sec}\:\theta\:\:,\:\:{dx}=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta\:{d}\theta \\ $$$$\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}\:}=\:\mathrm{3}\:\sqrt{\mathrm{sec}^{\mathrm{2}} \:\theta−\mathrm{1}\:}=\:\mathrm{3}\:\mid\mathrm{tan}\:\theta\mid=\:\mathrm{3}\:\mathrm{tan}\:\theta \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\left(\mathrm{1}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sec}\:\theta\right)\mathrm{sec}\:\theta\:{d}\theta \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{tan}\:\theta\:+\:{c} \\ $$$$\:{Since}\:\mathrm{sec}\:\theta=\:\frac{\sqrt{\mathrm{2}}\:\left({x}−\mathrm{1}\right)}{\mathrm{3}}\:\Rightarrow\:\mathrm{tan}\:\theta=\:\frac{\sqrt{\mathrm{2}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}}}{\mathrm{3}} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}\:\left({x}−\mathrm{1}\right)\:+\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}\:}\:}{\mathrm{3}}\mid+\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}\:}\:+\:{c} \\ $$$$ \\ $$

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