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Question Number 58220 by maxmathsup by imad last updated on 20/Apr/19
find∫dx(x2+x)−x2+2x+3
Answered by tanmay last updated on 20/Apr/19
−x2−x+3x+3−x(x+1)+3(x+1)(x+1)(3−x)t2=x+1dx=2tdt∫dxx(x+1)(x+1)(3−x)∫2tdt(t2−1)t2t2(3−t2+1)∫2dt(t2−1)t24−t22∫t2−(t2−1)t2(t2−1)4−t2dt2∫dt(t2−1)4−t2−2∫dtt24−t22I1−2I2I1=∫dt(t2−1)4−t2a2=4−t2t2−1→a2t2−a2=4−t2t2(a2+1)=a2+4t2=a2+4a2+12tdt=(a2+1)(2a)−(a2+4)(2a)(a2+1)2da2tdt=2a3+2a−2a3−8a(a2+1)2datdt=−3ada(a2+1)2∫−3adaa2+4a2+1×(a2+1)2×1(a2+4a2+1−1)4−(a2+4a2+1)∫−3adaa2+4×a2+1(a2+1)2×(a2+1)34a2+4−a2−4a2+1∫−3adaa2+4×133a−13∫daa2+4→{−13ln(a+a2+4)}so2I1=−23ln(4−t2t2−1+4−t2t2−1+4)=−23ln(4−t2+t3t2−1)←valueofI1lengthyproblemwaitforI2I2=∫dtt24−t2t=2sinθ∫2cosθdθ4sin2θ×2cosθ=12∫cosec2θdθ=−cotθ2=−12(4−t2t)←valueofI2socompleteanswdris2I1−2I22×−23ln(4−t2+t3t2−1)−2×−12(4−t2t)+ct2=x+1=−43ln(3−x+3(x+1)x)+(3−xx+1)+cplscheck...
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