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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
evaluate∫(t2+1+34t)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
letputI=∫t2+34t+1dtI=12∫4t2+3t+4dtbut4t2+3t+4=(2t)2+2(2t).34+916−916=(2t+34)2−916changement2t+34=34ch(x)giveI=12∫916(ch2x−1)38sh(x)dt=316.34∫sh2xdt=964∫ch(2x)−12dt=9128x−912812sh(2x)+λbutwehavech(x)=43(2t+34)=8t3+1⇒x=argch(8t3+1)=ln(8t3+1+(8t3+1)2−1)sh(x)=ch2x−1=(8t3−1)2−1andsh(2x)=2sh(x)ch(x)⇒I=9128ln(8t3+1+(8t3+1)2−1)−9128(8t3+1)(8t3−1)2−1+λ
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(x)=(8t3+1)2−1andI=9128ln(8t3+1+(8t3+1)2−1)−9128(8t3+1)(8t3+1)2−1+λ
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 .
∫(t+38)2+(558)2dt=(8t+316)t2+3t4+1+55128ln∣t+38+t2+3t4+1∣+c.

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