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Question Number 175436 by mathlove last updated on 30/Aug/22

∫((2x^2 +3)/( (√(3x+2))))=?

$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}=? \\ $$

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

Synthax error! haha!

$$\mathrm{Synthax}\:\mathrm{error}!\:\mathrm{haha}! \\ $$

Commented by a.lgnaoui last updated on 30/Aug/22

posons t=(√(3x+2)) x=((t^2 −2)/3) dx=2tdt ∫((2x^2 +3)/( (√(3x+2))))dx=∫((2(t^2 −2)^2 +27)/(9t))2tdt=∫((4(t^2 −2)^2 +54)/9)dt =∫[(4/9)(t^4 −4t^2 )+((70)/9)]dt] =(4/(45))t^5 −((16)/(27))t^3 +((70)/9)t+C ∫((2x^2 +3)/( (√(3x+2))))=[(4/(45))(3x+2)^2 −((16)/(27))(3x+2)+((70)/9)](√(3x+2)) +C

$${posons}\:{t}=\sqrt{\mathrm{3}{x}+\mathrm{2}}\:\:\:\:\:{x}=\frac{{t}^{\mathrm{2}} −\mathrm{2}}{\mathrm{3}}\:\:{dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx}=\int\frac{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\mathrm{27}}{\mathrm{9}{t}}\mathrm{2}{tdt}=\int\frac{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\mathrm{54}}{\mathrm{9}}{dt} \\ $$$$\left.=\int\left[\frac{\mathrm{4}}{\mathrm{9}}\left({t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} \right)+\frac{\mathrm{70}}{\mathrm{9}}\right]{dt}\right] \\ $$$$=\frac{\mathrm{4}}{\mathrm{45}}{t}^{\mathrm{5}} −\frac{\mathrm{16}}{\mathrm{27}}{t}^{\mathrm{3}} +\frac{\mathrm{70}}{\mathrm{9}}{t}+{C} \\ $$$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}=\left[\frac{\mathrm{4}}{\mathrm{45}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{27}}\left(\mathrm{3}{x}+\mathrm{2}\right)+\frac{\mathrm{70}}{\mathrm{9}}\right]\sqrt{\mathrm{3}{x}+\mathrm{2}}\:+{C}\:\:\:\:\:\:\: \\ $$

Answered by Ar Brandon last updated on 30/Aug/22

I=∫((2x^2 +3)/( (√(3x+2))))dx=∫((2x^2 )/( (√(3x+2))))dx+∫(3/( (√(3x+2))))dx =(2/9)∫(((3x+2−2)^2 )/( (√(3x+2))))dx+2(√(3x+2)) =(2/9)∫(((3x+2)^2 −4(3x+2)+4)/( (√(3x+2))))dx+2(√(3x+2)) =(2/9)∫((3x+2)^(3/2) −4(3x+2)^(1/2) +4(3x+2)^(−(1/2)) )dx+2(√(3x+2)) =(2/(27))((2/5)(3x+2)^(5/2) −(8/3)(3x+2)^(3/2) +8(3x+2)^(1/2) )+2(3x+2)^(1/2) +C =(4/(135))(3x+2)^(5/2) −((16)/(81))(3x+2)^(3/2) +((70)/(27))(3x+2)^(1/2) +C

$${I}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx}+\int\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{9}}\int\frac{\left(\mathrm{3}{x}+\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx}+\mathrm{2}\sqrt{\mathrm{3}{x}+\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{9}}\int\frac{\left(\mathrm{3}{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}{x}+\mathrm{2}\right)+\mathrm{4}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx}+\mathrm{2}\sqrt{\mathrm{3}{x}+\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{9}}\int\left(\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{4}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{4}\left(\mathrm{3}{x}+\mathrm{2}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right){dx}+\mathrm{2}\sqrt{\mathrm{3}{x}+\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{27}}\left(\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} −\frac{\mathrm{8}}{\mathrm{3}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{8}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)+\mathrm{2}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{C} \\ $$$$\:\:=\frac{\mathrm{4}}{\mathrm{135}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} −\frac{\mathrm{16}}{\mathrm{81}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{70}}{\mathrm{27}}\left(\mathrm{3}{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{C} \\ $$

Answered by Ar Brandon last updated on 30/Aug/22

I=∫((2x^2 +3)/( (√(3x+2))))dx, t=3x+2 ⇒x=((t−2)/3) =(1/3)∫((2(((t−2)/3))^2 +3)/( (√t)))dt=(1/(27))∫((2t^2 −8t+35)/( (√t)))dt =(1/(27))((4/5)t^(5/2) −((16)/3)t^(3/2) +70t^(1/2) )+C

$${I}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{2}}}{dx},\:{t}=\mathrm{3}{x}+\mathrm{2}\:\Rightarrow{x}=\frac{{t}−\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}\left(\frac{{t}−\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{{t}}}{dt}=\frac{\mathrm{1}}{\mathrm{27}}\int\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{35}}{\:\sqrt{{t}}}{dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{\mathrm{4}}{\mathrm{5}}{t}^{\frac{\mathrm{5}}{\mathrm{2}}} −\frac{\mathrm{16}}{\mathrm{3}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{70}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)+{C} \\ $$

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