Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 42580 by Raj Singh last updated on 28/Aug/18

Commented by maxmathsup by imad last updated on 28/Aug/18

changement (√(x+1))=t give x=t^2 −1 ⇒ ∫ ((x+(√(x+1)))/(x+2))dx = ∫ ((t^2 −1 +t)/(t^2 −1+2))(2t)dt = ∫ ((2t^3 +2t^2 −2t)/(t^2 +1))dt = 2 ∫ ((t(t^2 +1−1) +t^2 −t)/(t^2 +1))dt =2 ∫ ((t(t^2 +1) +t^2 −2t)/(t^2 +1))dt =2 ∫ t dt +2 ∫ ((t^2 +1 −2t−1)/(t^2 +1))dt = t^2 + 2t −2 ∫ ((2t+1)/(t^2 +1))dt =t^2 +2t −2 ln(t^2 +1) −2 arctan(t) +c = x+1 +2(√(x+1)) −2ln( x+2) −2 arctan((√(x+1))) +c

changementx+1=tgivex=t2−1⇒∫x+x+1x+2dx=∫t2−1+tt2−1+2(2t)dt=∫2t3+2t2−2tt2+1dt=2∫t(t2+1−1)+t2−tt2+1dt=2∫t(t2+1)+t2−2tt2+1dt=2∫tdt+2∫t2+1−2t−1t2+1dt=t2+2t−2∫2t+1t2+1dt=t2+2t−2ln(t2+1)−2arctan(t)+c=x+1+2x+1−2ln(x+2)−2arctan(x+1)+c

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

∫((x+2−2)/(x+2))dx+∫((√(x+1))/(x+2))dx ∫dx−2∫(dx/(x+2))+I_3 x−2ln(x+2)+I_3 t^2 =x+1 dx=2tdt ∫((t×2tdt)/(t^2 +1)) 2∫((t^2 +1−1)/(t^2 +1))dt 2∫dt−2∫(dt/(t^2 +1)) 2t−2tan^(−1) (t) 2((√(x+1)) )−2tan^(−1) ((√(x+1)) ) so ans is x−2ln(x+2)+2(√(x+1)) )−2tan^(−1) ((√(x+1)) )+c

∫x+2−2x+2dx+∫x+1x+2dx∫dx−2∫dxx+2+I3x−2ln(x+2)+I3t2=x+1dx=2tdt∫t×2tdtt2+12∫t2+1−1t2+1dt2∫dt−2∫dtt2+12t−2tan−1(t)2(x+1)−2tan−1(x+1)soansisx−2ln(x+2)+2x+1)−2tan−1(x+1)+c

Terms of Service

Privacy Policy

Contact: info@tinkutara.com