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Question Number 92323 by Power last updated on 06/May/20

Commented by Prithwish Sen 1 last updated on 06/May/20

split x+3 into (1/8)(8x+4)+(5/2)

$$\mathrm{split}\:\boldsymbol{\mathrm{x}}+\mathrm{3}\:\boldsymbol{\mathrm{into}}\:\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{8}\boldsymbol{\mathrm{x}}+\mathrm{4}\right)+\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by Power last updated on 06/May/20

solve sir pls

$$\mathrm{solve}\:\mathrm{sir}\:\mathrm{pls} \\ $$

Commented by Zainal Arifin last updated on 09/May/20

$$ \\ $$

Answered by niroj last updated on 06/May/20

  ∫  ((x+3)/(√(4x^2 +4x+5)))dx     =   (1/8)∫ ((8x+24)/(√(4x^2 +4x+5)))dx    = (1/8)∫ ((8x+4+20)/(√(4x^2 +4x+5)))dx    =  (1/8)∫ ((8x+4)/(√(4x^2 +4x+5)))dx+((20)/8)∫ (( dx)/(√(4x^2 +4x+5)))    = (1/4)(√(4x^2 +4x+5))  + (5/2)∫ (dx/(√((2x)^2 +2.2x.1+1+4)))+C    =((√(4x^2 +4x+5))/4) +(5/2)∫ (1/(√((2x+1)^2 +(2)^2 )))dx+C   = ((√(4x^2 +4x+5))/4) + (5/2).(1/2)[ log (2x+1+(√((2x+1)^2 +(2)^2 ))  )]+C   =  (1/4)(√(4x^2 +4x+5))   + (5/4)log (2x+1+(√(4x^2 +4x+5)) )+C //.

$$\:\:\int\:\:\frac{\boldsymbol{\mathrm{x}}+\mathrm{3}}{\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{5}}}\boldsymbol{\mathrm{dx}} \\ $$$$\:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\int\:\frac{\mathrm{8x}+\mathrm{24}}{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}}\mathrm{dx} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{8}}\int\:\frac{\mathrm{8x}+\mathrm{4}+\mathrm{20}}{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}}\mathrm{dx} \\ $$$$\:\:=\:\:\frac{\mathrm{1}}{\mathrm{8}}\int\:\frac{\mathrm{8x}+\mathrm{4}}{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}}\mathrm{dx}+\frac{\mathrm{20}}{\mathrm{8}}\int\:\frac{\:\mathrm{dx}}{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}\:\:+\:\frac{\mathrm{5}}{\mathrm{2}}\int\:\frac{\mathrm{dx}}{\sqrt{\left(\mathrm{2x}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{2x}.\mathrm{1}+\mathrm{1}+\mathrm{4}}}+{C} \\ $$$$\:\:=\frac{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}}{\mathrm{4}}\:+\frac{\mathrm{5}}{\mathrm{2}}\int\:\frac{\mathrm{1}}{\sqrt{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} }}\mathrm{dx}+{C} \\ $$$$\:=\:\frac{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}}{\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\left[\:\mathrm{log}\:\left(\mathrm{2x}+\mathrm{1}+\sqrt{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} }\:\:\right)\right]+\mathrm{C} \\ $$$$\:=\:\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}\:\:\:+\:\frac{\mathrm{5}}{\mathrm{4}}\mathrm{log}\:\left(\mathrm{2x}+\mathrm{1}+\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}\:\right)+\mathrm{C}\://. \\ $$

Commented by Power last updated on 06/May/20

thanks

$$\mathrm{thanks} \\ $$

Commented by niroj last updated on 06/May/20

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Commented by MJS last updated on 06/May/20

a minor mistake...

$$\mathrm{a}\:\mathrm{minor}\:\mathrm{mistake}... \\ $$

Commented by niroj last updated on 06/May/20

 thanks mr.mjs for your correction .

$$\:\mathrm{thanks}\:\mathrm{mr}.\mathrm{mjs}\:\mathrm{for}\:\mathrm{your}\:\mathrm{correction}\:. \\ $$

Answered by MJS last updated on 06/May/20

∫((x+3)/(√(4x^2 +4x+5)))dx=       [t=((2x+1+(√(4x^2 +4x+5)))/2) → dx=((√(4x^2 +4x+5))/(2x+1+(√(4x^2 +4x+5))))dt]  =∫((t^2 +5t−1)/(4t^2 ))dt=(1/4)∫dt+(5/4)∫(dt/t)−(1/4)∫(dt/t^2 )=  =(1/4)t+(5/4)ln t +(1/(4t))=  =((√(4x^2 +4x+5))/4)+(5/4)ln (2x+1+(√(4x^2 +4x+5))) +C

$$\int\frac{{x}+\mathrm{3}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}}{\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}}{dt}\right] \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{5}{t}−\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{4}}\int{dt}+\frac{\mathrm{5}}{\mathrm{4}}\int\frac{{dt}}{{t}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{t}+\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ln}\:{t}\:+\frac{\mathrm{1}}{\mathrm{4}{t}}= \\ $$$$=\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}\right)\:+{C} \\ $$

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