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Question Number 86302 by john santu last updated on 28/Mar/20
∫dx(x2+1)x2+4=?
Commented by abdomathmax last updated on 28/Mar/20
I=∫dx(x2+1)x2+4changementx=2sh(t)giveI=∫2ch(t)dt(1+4sh2t)2ch(t)=∫dt1+4×ch(2t)−12=∫2dt2+4ch(2t)−4=∫dt2ch(2t)−1=∫dt2×e2t+e−2t2−1=∫dte2t+e−2t−1=et=z∫dzz(z2+z−2−1)=∫dzz3−z−1−z=∫zdzz4−1−z2=∫zdzz4−z2−1letdecomposeF(z)=zz4−z2−1z4−z2−1=0⇒t2−t−1=0(t=z2)Δ=1+4=5⇒t1=1+5andt2=1−5⇒F(z)=z(z2−t1)(z2−t2)=z25(1z2−t1−1z2−t2)⇒∫F(z)dz=125ln(z2−t1z2−t2)+C=125ln(e2t−t1e2t−t2)+Ct=argsh(x2)=ln(x2+1+x24)⇒e2t=(x2+1+x24)2⇒I=125ln((x2+1+x22)2−1−5(x2+1+x24)2−1+5)+C
Answered by TANMAY PANACEA. last updated on 28/Mar/20
x=1t→dx=−dtt2∫−dtt2(1t2+1)1t2+4∫−tdt(1+t2)1+4t2t2=k→dk=2tdt∫−dk2(1+k)1+4k−12∫dk(1+k)1+4k1+4k=p24dk=2pdp−12∫pdp2×(1+p2−14)p=−14∫4dp(p2+3)=−13tan−1(p3)+c=−13tan−1(1+4k3)+c=−13tan−1(1+4t23)+c=−13tan−1(1+4x23)+c
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