Transcription of Integration by Partial Fractions
1 Joe FosterIntegration by Partial FractionsSummary: Method of Partial Fractions whenf(x)g(x)is proper (degf(x)<degg(x))1. Letx rbe a linear factor ofg(x). Suppose that (x r)mis the highest power ofx rthat dividesg(x). Then, tothis factor, assign the sum of thempartial Fractions :A1(x r)+A2(x r)2+A3(x r)3+ +Am(x r) this for each distinct linear factor ofg(x).2. Letx2+px+qbe an irreducible quadratic factor ofg(x) so thatx2+px+qhas no real roots. Suppose that(x2+px+q)nis the highest power of this factor that dividesg(x). Then, to this factor, assign the sum of thenpartial Fractions :B1x+C1(x2+px+q)+B2x+C2(x2+px+q )2+B3x+C3(x2+px+q)3+ +Bnx+Cn(x2+px+q) this for each distinct quadratic factor ofg(x).3. Continue with this process with all irreducible factors,and all powers.
2 The key things to remember are(i) One fraction for each power of the irreducible factor that appears(ii) The degree of the numerator should be one less than the degree of the denominator4. Set the original fractionf(x)g(x)equal to the sum of all these Partial Fractions . Clear the resulting equation of fractionsand arrange the terms in decreasing powers Solved for the undetermined coefficients by either strategically plugging in values or comparing coefficients of 1 Compute x+ 14(x+ 5) (x+ 2) first step is to decomposex+ 14(x+ 5) (x+ 2)asx+ 14(x+ 5) (x+ 2)=?x+ 5+?x+ is no indicator of what the numerators should be, so there is work to be done to find them. If we let the numeratorsbe variables, we can use algebra to solve. That is, we want to find constantsAandBthat make the equation below truefor allx6= 5, + 14(x+ 5) (x+ 2)=Ax+ 5+Bx+ solve forAandBby cross multiplying and equating the + 14(x+ 5) (x+ 2)=Ax+ 5+Bx+ 2=A(x+ 2) +B(x+ 5)(x+ 5) (x+ 2)= x+ 14 =A(x+ 2) +B(x+ 5)=Ax+ 2A+Bx+ 5B= (A+B)x+ 2A+ 5 BPage 1 of 4 MATH 142 - Integration by Partial FractionsJoe FosterThis leaves us with the system of equationsA+B= 12A+ 5B= the first we obtainB= 1 A.
3 Substituting this into the second gives14 = 2A+ 5B= 2A+ 5 (1 A) = 2A+ 5 5A= 5 3A= 9 = 3A= A= 3= B= 4So, x+ 14(x+ 5) (x+ 2)dx= 3x+ 5+4x+ 2dx= 3 ln|x+ 5|+ 4 ln|x+ 2|+CThis method isn t a new way to integrate. This method is just an exercise in algebraic manipulation to rearrange aseemingly complicated integral to turn it into an integral that can be done using the methods we are familiar with. Let snow see an example of when there is a repeated irreducible factor on the 2 Find 5x 2(x+ 3) , there are not two different linear factors in the denominator. This CANNOT be expressed in the form5x 2(x+ 3)2=5x 2(x+ 3)(x+ 3)6=Ax+ 3+Bx+ 3=A+Bx+ , it can be expressed in the form:5x 2(x+ 3)2=Ax+ 3+B(x+ 3) 2(x+ 3)2=Ax+ 3+B(x+ 3)2=A(x+ 3) +B(x+ 3)2= 5x 2 =A(x+ 3) +Bx= 3 :5( 3) 2 =A(( 3) + 3) +B= 17 = 0 +B= 17 =B5x 2 =A(x+ 3) 17=Ax+ 3A 17= 5x=Ax= 5 =A 5x 2(x+ 3)2dx= 5x+ 3 17(x+ 3)2dx= 5 ln|x+ 3|+17x+ 3+CThe final thing we should look at is the case when there is an irreducible polynomial of degree higher than 1 on thedenominator of the rational 2 of 4 MATH 142 - Integration by Partial FractionsJoe FosterExample 3 Compute 2x+ 4(x2+ 1)(x 1) process follows as before.
4 The most common mistake here is to not choose the right numerator for the term with thex2+ 1 on the denominator. The term of the numerator should have degree 1 less than the denominator - so this termshould have anAx+Bon the numerator. Let s see: 2x+ 4(x2+ 1)(x 1)=Ax+Bx2+ 1+Cx 1=(Ax+B)(x 1) +C(x2+ 1)(x2+ 1)(x 1)= 2x+ 4 = (Ax+B)(x 1) +C(x2+ 1).x= 1 : 2(1) + 4 = (Ax+B)(1 1) +C((1)2+ 1)= 2 = 0 + 2C= 1 =CSo now we have, 2x+ 4 = (Ax+B)(x 1) + (x2+ 1)= x2 2x+ 3 = (Ax+B)(x 1)= (x 1)(x+ 3) = (Ax+B)(x 1)= (x+ 3) =Ax+B= 1 =A 3 = our integral becomes, 2x+ 4(x2+ 1)(x 1)dx= x 3x2+ 1+1x 1dx= xx2+ 1dx 3 1x2+ 1dx+ 1x 1dx= 12ln(x2+ 1) 3 tan 1(x) + ln|x 1|+CWhere the first two integrals are solved with au-substitution and trigonometric substitution, ProblemsTry some of the problems below.
5 If you get stuck, don t worry!There are hints on the next page! But do try withoutlooking at them first, chances are you won t get hints on your x 9(x+ 5)(x 2)dx2. 1(t+ 4)(t 1)dt3. 321x2 1dx4. 10x 1x2+ 3x+ 2dx5. 214y2 7y 12y(y+ 2)(y 3)dy6. x2+ 2x 1x3 xdxPage 3 of 4 MATH 142 - Integration by Partial FractionsJoe Foster7. 1(x+ 5)2(x 1)dx8. x2 5x+ 16(2x+ 1)(x 2)2dx9. 1s2(s 1)2ds10. 5x2+ 3x 2x3+ 2x2dx11. x2 x+ 6x3+ 3xdx12. 10(x 1)(x2+ 9)dxChallenge ProblemsBelow are some harder problems that require a little more thinking/algebraic manipulation to make the 43x3 2x2 4x3 2x2dx2. x3 4x 10x2 x 6dx3. x3+x2+ 2x+ 1(x2+ 1)(x2+ 2)dx4. x2 2x 1(x 1)2(x2+ 1)dx5. x+ 4x2+ 2x+ 5dx6. x2+x+ 1(x2+ 1)2dxAnswers to Practice Problems1. 2 ln|x+ 5| ln|x 2|+ (ln|t 1| ln|t+ 4|) + (32)4.
6 Ln(2732) (83)6. ln|x|+ ln|x 1| ln|x+ 1|+ (6x+ 5+ ln|x 1| ln|x+ 5|)+ (3 ln|2x+ 1| 2 ln|x 2| 4x 2)+C9. 2 ln|s| 1s 2 ln|s 1| 1s 1+C10. 2 ln|x|+1x+ 3 ln|x+ 2|+ (4 ln|x| ln(x2+ 3))+ (6 ln|x 1| 3 ln(x2+ 9) 2 tan 1(x3))+CAnswers to Challenge (7 + 6 ln(23)) (x2+ 2x+ 2 ln|x 3|+ 4 ln|x+ 2|)+ (ln(x2+ 1) + 2 tan 1( 2x2))+ (2 ln|x 1|+2x 1 ln(x2+ 1) + 2 tan 1(x))+ (ln(x2+ 2x+ 5) + 3 tan 1(x+ 12))+ (2 tan 1(x) 1x2+ 1)+CPage 4 of 4
