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Question Number 87995 by abdomathmax last updated on 07/Apr/20
find ∫ (dx/((x^2 −1)^2 (x^2 +2)))
find∫dx(x2−1)2(x2+2)
Answered by redmiiuser last updated on 08/Apr/20
(1/(x^2 −1))=t (dx/((x^2 −1)^(2 ) ))=−dt x^2 +2=(1/t)+3=((1+3t)/t) ∫((−t.dt)/(1+3t)) 1+3t=z 3.dt=dz ((−1)/3)∫(((z−1).dz)/(3.z)) =((−1)/9)[∫dz−∫(dz/z)] =((−1)/9)[z−lnz]+c =(1/9)[ln z−z]+c =(1/9)[ln (((x^2 +2)/(x^2 −1)))−((x^2 +2)/(x^2 −1))]+c
1x2−1=tdx(x2−1)2=−dtx2+2=1t+3=1+3tt∫−t.dt1+3t1+3t=z3.dt=dz−13∫(z−1).dz3.z=−19[∫dz−∫dzz]=−19[z−lnz]+c=19[lnz−z]+c=19[ln(x2+2x2−1)−x2+2x2−1]+c

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