By solving these equations we obtain a 1 b 0 c 2 d 0 e 1 f g h 0 thus I x 4 2 x

# By solving these equations we obtain a 1 b 0 c 2 d 0

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By solving these equations we obtain:a=1, b= 0, c=2, d= 0, e=1, f=g=h= 0,thusI=x42x21(x+ 1)(x2+x+ 1)2+C.Problems 3.5. (a) Find the integralI=5x3+9x222x8x34xdx. Use the Partial Fraction Decomposition.(b) Find the integralI=x3+1x(x1)3dx. Use the method of Ostrogradsky.(c) Find the integralI=x3+3(x+1)(x2+1)dx.Solution.Problem??(a):We have that5x3+ 9x222x8x34x=5 +9x22x8x34x= 5 +9x22x8x(x2)(x+ 2)=52x+5x2+6x+ 2.22
Thus5x3+ 9x222x8x34xdx=∫ (52x+5x2+6x+ 2)dx=5x2 ln|x|+ 5 ln|x2|+ 6 ln|x+ 2|+C=5x+ ln(x2)5(x+ 2)6x2+CProblem??(b):We look for the polynomialsP1(x) =Ax+BandP2(x) =Cx+Dsuch thatx3+ 1x(x1)3dx=P1(x)Q1(x)+P2(x)Q2(x)dx,whereQ1(x) = (x1)2andQ2(x) =x(x1).We putH(x) =Q1(x)Q2(x)Q1(x)= 2xand we find thecoeﬃcientsA, B, C, Dfrom the equationP1(x)Q2(x)P1(x)H(x) +P2(x)Q1(x) =x3+ 1,which can be written asAx(x1)(Ax+B)2x+ (Cx+D)(x1)2=x3+ 1.After simplification, we getCx3+ (A+D2C)x2+ (A2B2D+C)x+D=x3+ 1,soA=1,B= 0,C= 1, andD= 1. Consequently,x3+ 1x(x1)3dx=x(x1)2+x+ 1x(x1)dx=x(x1)2+∫ (1x+2x1)dx=x(x1)2+ ln(x1)2x+C.Problem??(c):We have the follwing partial fraction decomposition:x3+ 3(x+ 1)(x2+ 1)= 1 +x2x+ 2(x+ 1)(x2+ 1)= 1 +1x+ 1+12xx2+ 1thusx3+ 3(x+ 1)(x2+ 1)dx=∫ (1 +1x+ 12xx2+ 1+1x2+ 1)dx=x+ ln|x+ 1| −ln|x2+ 1|+ arctan(x) +C=x+ lnx+ 1x2+ 1+ arctan(x) +C.23
4Integration of Certain Irrational ExpressionsWe begin with integrals of typeI=R(x,mαx+βγx+δ)dx,wheremis a natural number andRis a rational function of two variables, i.e.R(x, y) =ajixiyjblsxlys,where the sums standing in the numerator and the denominator are finite.We apply the substitutiont=ω(x) =mαx+βγx+δ,i.e.tm=αx+βγx+δ.Therefore,x=φ(t) =δtmβαγtm,and we obtain thatI=R(x,mαx+βγx+δ)dx=R(φ(t), t)φ(t)dt.Sinceφ(t),φ(t) andR(x, t) are rational functions, the integrantR(φ(t), t)φ(t) is also a rational func-tion, therefore the computation of the integralIis reduced to the computation of a rational integralR(φ(t), t)φ(t)dt.It is clear that the established method of integration of rational functions can beeffectively applied as long as we are able to representR(φ(t), t)φ(t) in a formP(t)Q(t), whereQ(t) can befactorized, i.e. we are able to compute its roots and irreducible quadratic factors.Example 8.We will compute the following integrals(a)I=x+1+2(x+1)2x+1dx. In this example the rational functionαx+βγx+δis reduced tox+ 1 andm= 2,therefore we apply the substitutiont=x+ 1, i.e.t2=x+ 1, thusdx= 2tdtand we obtainI=x+ 1 + 2(x+ 1)2x+ 1dx= 2t+ 2t31dt=∫ (2t12t+ 2t2+t+ 1)dt= ln(t1)2t2+t+ 123arctan2t+ 13+C= ln(