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x-x-2-25-x-dx-




Question Number 183806 by cortano1 last updated on 30/Dec/22
  ∫ ((√(x+(√(x^2 +25))))/x) dx =?
x+x2+25xdx=?
Answered by Ar Brandon last updated on 30/Dec/22
I=∫((√(x+(√(x^2 +25))))/x)dx , x=5shϑ    =∫((√(5shϑ+5chϑ))/(5shϑ))(5chϑdϑ)=(√5)∫(((e^(2ϑ) +1)/(e^(2ϑ) −1)))e^(ϑ/2) dϑ, ϑ=2φ    =2(√5)∫(((e^(4φ) +1)/(e^(4φ) −1)))e^φ dφ=2(√5)∫(((t^4 +1)/(t^4 −1)))dt=2(√5)∫(1+(2/(t^4 −1)))dt, t=e^φ     =2(√5)∫(1+(1/(t^2 −1))−(1/(t^2 +1)))dt=2(√5)(t+(1/2)ln∣((t−1)/(t+1))∣−arctan(t))+C    =2(√5)(e^(ϑ/2) +(1/2)ln∣e^(ϑ/2) −1∣−(1/2)ln∣e^(ϑ/2) +1∣−arctan(e^(ϑ/2) ))+C    =2(√5)(√(x+(√(x^2 +25))))+(√5)ln∣(((√(x+(√(x^2 +25))))−1)/( (√(x+(√(x^2 +25))))+1))∣−2(√5)arctan((√(x+(√(x^2 +25)))))+C
I=x+x2+25xdx,x=5shϑ=5shϑ+5chϑ5shϑ(5chϑdϑ)=5(e2ϑ+1e2ϑ1)eϑ2dϑ,ϑ=2ϕ=25(e4ϕ+1e4ϕ1)eϕdϕ=25(t4+1t41)dt=25(1+2t41)dt,t=eϕ=25(1+1t211t2+1)dt=25(t+12lnt1t+1arctan(t))+C=25(eϑ2+12lneϑ2112lneϑ2+1arctan(eϑ2))+C=25x+x2+25+5lnx+x2+251x+x2+25+125arctan(x+x2+25)+C

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