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1-1-x-2-2x-dx-

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Question Number 181735 by Mastermind last updated on 29/Nov/22
∫(1/( (√(1+x^2 ))−2x))dx
∫11+x2−2xdx
Answered by Gamil last updated on 29/Nov/22
Answered by MJS_new last updated on 30/Nov/22
∫(dx/(−2x+(√(x^2 +1))))= [t=x+(√(x^2 +1)) → dx=((t^2 +1)/(2t^2 ))dt] =−∫((t^2 +1)/(t(t^2 −3)))dt= =∫((1/(3t))−(2/(3(t−(√3))))−(2/(3(t+(√3)))))dt= =(1/3)ln t −(2/3)(ln (t−(√3)) +ln (t+(√3)))= =(1/3)ln (t/((t^2 −3)^2 )) = =(1/3)ln ((x^2 +1)^(3/2) −x(x^2 −3)) −(2/3)ln ∣3x^2 −1∣ +C
∫dx−2x+x2+1=[t=x+x2+1→dx=t2+12t2dt]=−∫t2+1t(t2−3)dt==∫(13t−23(t−3)−23(t+3))dt==13lnt−23(ln(t−3)+ln(t+3))==13lnt(t2−3)2==13ln((x2+1)3/2−x(x2−3))−23ln∣3x2−1∣+C

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