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Question Number 35412 by mondodotto@gmail.com last updated on 18/May/18

evaluate ∫(√((t^2 +1+(3/4)t))) dt

$$\boldsymbol{\mathrm{evaluate}}\:\int\sqrt{\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{\mathrm{t}}\right)}\:\boldsymbol{\mathrm{dt}} \\ $$

Commented by prof Abdo imad last updated on 19/May/18

let put I = ∫ (√(t^2 +(3/4)t +1)) dt I = (1/2) ∫ (√(4t^2 +3t+4)) dt but 4t^2 +3t +4 =(2t)^2 +2 (2t).(3/4) +(9/(16)) −(9/(16)) =(2t +(3/4))^2 −(9/(16)) changement 2t+(3/4) =(3/4)ch(x) give I = (1/2) ∫ (√((9/(16))(ch^2 x−1))) (3/8)sh(x)dt =(3/(16)) .(3/4) ∫ sh^2 x dt = (9/(64)) ∫ ((ch(2x)−1)/2)dt = (9/(128)) x −(9/(128)) (1/2)sh(2x) +λ but we have ch(x)=(4/3)(2t+(3/4)) =((8t)/3) +1 ⇒x=argch(((8t)/3) +1) =ln( ((8t)/3) +1 +(√((((8t)/3)+1)^2 −1))) sh(x)=(√(ch^2 x−1)) = (√((((8t)/3)−1)^2 −1)) and sh(2x)=2sh(x)ch(x)⇒ I =(9/(128))ln( ((8t)/3) +1+(√((((8t)/3)+1)^2 −1))) −(9/(128)) (((8t)/3)+1)(√((((8t)/3)−1)^2 −1)) +λ

$${let}\:{put}\:{I}\:=\:\int\:\:\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}{t}\:+\mathrm{1}}\:{dt} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{3}{t}+\mathrm{4}}\:{dt}\:\:{but} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{4}\:=\left(\mathrm{2}{t}\right)^{\mathrm{2}} \:+\mathrm{2}\:\left(\mathrm{2}{t}\right).\frac{\mathrm{3}}{\mathrm{4}}\:\:+\frac{\mathrm{9}}{\mathrm{16}}\:−\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$=\left(\mathrm{2}{t}\:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:−\frac{\mathrm{9}}{\mathrm{16}}\:\:{changement}\:\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{4}}{ch}\left({x}\right) \\ $$$${give}\:{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\sqrt{\frac{\mathrm{9}}{\mathrm{16}}\left({ch}^{\mathrm{2}} {x}−\mathrm{1}\right)}\:\:\frac{\mathrm{3}}{\mathrm{8}}{sh}\left({x}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\:.\frac{\mathrm{3}}{\mathrm{4}}\:\int\:\:{sh}^{\mathrm{2}} {x}\:{dt} \\ $$$$=\:\frac{\mathrm{9}}{\mathrm{64}}\:\int\:\:\frac{{ch}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$=\:\frac{\mathrm{9}}{\mathrm{128}}\:{x}\:−\frac{\mathrm{9}}{\mathrm{128}}\:\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{x}\right)\:+\lambda\:\:\:{but}\:{we}\:{have} \\ $$$${ch}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\frac{\mathrm{8}{t}}{\mathrm{3}}\:+\mathrm{1}\:\Rightarrow{x}={argch}\left(\frac{\mathrm{8}{t}}{\mathrm{3}}\:+\mathrm{1}\right) \\ $$$$={ln}\left(\:\frac{\mathrm{8}{t}}{\mathrm{3}}\:+\mathrm{1}\:+\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}}\right) \\ $$$${sh}\left({x}\right)=\sqrt{{ch}^{\mathrm{2}} {x}−\mathrm{1}}\:=\:\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}}\:\:\:{and} \\ $$$${sh}\left(\mathrm{2}{x}\right)=\mathrm{2}{sh}\left({x}\right){ch}\left({x}\right)\Rightarrow \\ $$$${I}\:=\frac{\mathrm{9}}{\mathrm{128}}{ln}\left(\:\frac{\mathrm{8}{t}}{\mathrm{3}}\:+\mathrm{1}+\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}}\right) \\ $$$$−\frac{\mathrm{9}}{\mathrm{128}}\:\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\:\:\:+\lambda \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 19/May/18

sh(x)= (√((((8t)/3)+1)^2 −1)) and I = (9/(128))ln(((8t)/3)+1 +(√((((8t)/3)+1)^2 −1))) −(9/(128))(((8t)/3)+1)(√((((8t)/3)+1)^2 −1)) +λ

$${sh}\left({x}\right)=\:\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}}\:\:\:{and} \\ $$$${I}\:=\:\frac{\mathrm{9}}{\mathrm{128}}{ln}\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\:+\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$−\frac{\mathrm{9}}{\mathrm{128}}\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)\sqrt{\left(\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\:+\lambda \\ $$

Answered by ajfour last updated on 18/May/18

∫(√((t+(3/8))^2 +(((√(55))/8))^2 )) dt =(((8t+3)/(16)))(√(t^2 +((3t)/4)+1)) +((55)/(128))ln ∣t+(3/8)+(√(t^2 +((3t)/4)+1)) ∣+c .

$$\int\sqrt{\left({t}+\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{55}}}{\mathrm{8}}\right)^{\mathrm{2}} }\:{dt} \\ $$$$=\left(\frac{\mathrm{8}{t}+\mathrm{3}}{\mathrm{16}}\right)\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{3}{t}}{\mathrm{4}}+\mathrm{1}}\:+\frac{\mathrm{55}}{\mathrm{128}}\mathrm{ln}\:\mid{t}+\frac{\mathrm{3}}{\mathrm{8}}+\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{3}{t}}{\mathrm{4}}+\mathrm{1}}\:\mid+{c}\:. \\ $$

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