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Question Number 156729 by MathSh last updated on 14/Oct/21

Answered by MJS_new last updated on 15/Oct/21

∫((x+2)/((x^2 +3x+3)(√(x+1))))dx= [t=(√(x+1)) → dx=2(√(x+1))dt] =2∫((t^2 +1)/(t^4 +t^2 +1))dt=∫(dt/(t^2 −t+1))+∫(dt/(t^2 +t+1))= =((2(√3))/3)arctan (((√3)(2t−1))/3) +((2(√3))/3)arctan (((√3)(2t+1))/3) = =((2(√3))/3)(arctan (((√3)(2(√(x+1))−1))/3) +arctan (((√3)(2(√(x+1))+1))/3))+C

∫x+2(x2+3x+3)x+1dx=[t=x+1→dx=2x+1dt]=2∫t2+1t4+t2+1dt=∫dtt2−t+1+∫dtt2+t+1==233arctan3(2t−1)3+233arctan3(2t+1)3==233(arctan3(2x+1−1)3+arctan3(2x+1+1)3)+C

Commented by MathSh last updated on 15/Oct/21

Very nice dear Ser thankyou

VerynicedearSerthankyou

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