7.4 Integration by Partial Fractions

1 Integration by Partial FractionsThe method of Partial Fractions is used to integraterational is, we want to compute P(x)Q(x)dxwhereP,Qare reduce1the integrand to the formS(x) +R(x)Q(x)where R< we write the integrand as a polynomial plus a rational function7x+2whose denom-inator has higher degreee than its numerator. Thankfully, this expression can be easily integratedusing +3x+2=x(x+2) 2x+3x+2=x+ 2(x+2) +4+3x+2=x 2+7x+2= x2+3x+2dx= x 2+7x+2dx=12x2 2x+7 ln|x+2|+cWhat if Q 2?If the denominatorQ(x)is quadratic or has higher degree, we need another that R< Q. Then the rational functionR(x)Q(x)can be written as a sum of Fractions of theformA(ax+b)mAx+B(ax2+bx+c)nwhere A,B,a,b,c are constants and m,n are positive such as the above can all be integrated using either logarithms or trigonometric a little experimenting, you should be convinced that3x2+2x+3x3+x=3x+21+x2It follows that 3x2+2x+3x3+xdx=3 ln|x|+2 tan 1x+cThe burning question ishowto find the expressions in the Therorem.

2 The approach depends onthe form of the denominatorQ(x).1By Long Division or some other Torture..1 Case 1: Distinct Linear FactorsSuppose that our denominator can be factorized completely intodistinctlinear factors. That isQ(x) = (x a1)(x a2) (x an)where the valuesa1, .. ,anare all such a Q, there exist constants A1, .. ,Ansuch thatR(x)Q(x)=n i=1 Aix ai=A1x a1+ +Anx an( )whence the integral can be easily computed term-by-term: R(x)Q(x)dx=n i=1 Aix aidx=n i=1 Ailn|x ai|+cWe find the constantsAiby putting the right hand side of ( ) over the common denominatorQ(x)R(x)Q(x)=R(x)(x a1) (x an)=A1x a1+ +Anx anand comparing According to the Theorem, there exist constantsA,Bsuch thatx+8x2+x 2=x+8(x 1)(x+2)=Ax 1+Bx+2 Summing the right hand side, we obtainx+8(x 1)(x+2)=A(x+2) +B(x 1)(x 1)(x+2)Since the denominators are equal, it follows that the numerators are equal:x+8=A(x+2) +B(x 1)This is a relationship betweenA,Bwhich holdsfor all3x: every value ofxgives a valid rela-tionship betweenAandB.

3 Evaluating atx=1 andx= 2 gives two very simple expressions:x=1 :9=3A= A=3x= 2 :6= 3B= B= 2 Putting it all together, we have x+8x2+x 2dx= 3x 1 2x+2dx=3 ln|x 1| 2 ln|x+2|+c=ln|x 1|3|x+2|2+c2We assume for clarity that the leading term ofQ(x)isxn(coefficient 1). If not, absorb it into the numerator!3 You might worry that it doesn t whenx=1 orx= 2 because of the denominator. The fact fact that polynomials arecontinuouscombined withx+8=A(x+2) +B(x 1)everywhere else guarantees that we have equality We know that there exist constantsA,B,Csuch thatx2+2x3 x=x2+2x(x 1)(x+1)=Ax+Bx 1+Cx+1 Combining the right hand side yieldsx2+2=A(x 1)(x+1) +Bx(x+1) +Cx(x 1)Now evaluate atx=0, 1:x=0 :2= A= A= 1x=1 :3=2B= B=32x= 1 :3=2C= C=32It follows that x2+2x3 xdx= 2x+32(x 1)+32(x+1)dx= 2 ln|x|+32(ln|x 1|+ln|x+1|) +c=ln|x2 1|32x2+cCase 2: Repeated Linear FactorsSuppose that when we factorizeQ(x)we obtain a repeated linear factor.

4 That is, some term of theform(x a)mwherem 2. In a Partial Fractions decomposition, such a factor producesmseperatecontributions:A1x a+A2(x a)2+ +Am(x a)meach of which can be integrated normally. One way to remember this is to count the constants:(x a)mhas degreemand must therefore correspond tomdistinct 2x2(x 1)has a repeated factor ofxin the denominator. The single factor ofx 1 behaves exactlyas in Case 1. We therefore have constantsA,B,Csuch thatx 2x2(x 1)=Ax+Bx2+Cx 1 Combining the right hand side and cancelling the denominators yields4x 2=Ax(x 1) +B(x 1) +Cx2( )4Be careful: think about what each term is missing compared to the common are only two nice places at which to evaluate this expression:x=0 : 2= B= B=2x=1 : 1=CTo obtainAwe have choices.

5 Either evaluate ( ) at another value ofx, or compare example, it is easy to see that the coefficient ofx2on the right side of ( ) isA+C. This isclearly zero, since ther is nox2term on the left. We might write this ascoeff(x2):0=A+C= A= C=1 Putting it all together, we have x 2x2(x 1)dx= 1x+2x2 1x 1dx=ln|x||x 1| 2x+c2. Suppose we want to integratex3+3x+1(x+1)2(x 2)2. We have two repeated factors, whence thereexist constantsA,B,C,Dsuch thatx3+3x+1(x+1)2(x 2)2=Ax+1+B(x+1)2+Cx 2+D(x 2)2 Combining the right hand side and cancelling the denominators yieldsx3+3x+1=A(x+1)(x 2)2+B(x 2)2+C(x+1)2(x 2) +D(x+1)2We evaluate at the two nice places then compare some coefficients and evaluate atx=0:x=2 :15=9D= D=53x= 1 : 3=9B= B= 13coeff(x3):1=A+Cx=0 :1=4A+4B 2C+D= 2A C=13 The last two equations can be solved to obtainA=49andC=59.

6 The final integral is then x3+3x+1(x+1)2(x 2)2dx= 49(x+1) 13(x+1)2+59(x 2)+53(x 2)2dx=49ln|x+1|+13(x+1)+59ln|x 2| 53(x 2)+c=19ln|x+1|4|x 2|5+13(x+1) 53(x 2)+c4 Case 3: Quadratic FactorsSuppose that the denominatorQ(x)contains anirreducible quadraticterm: a term of the form5ax2+bx+cwhereb2 4ac<0 Each such factor generates a Partial fraction of the formAx+Bax2+bx+cwhich can be integrated using logarithms and/or tangent rational functionx2 x+2x3+4x=x2 x+2x(x2+4)contains the irreduciuble quadraticx2+4in its denominator. We therefore know that there exist constantsA,B,Csuch thatx2 x+2x3+4x=Ax+Bx+Cx2+4 Combining the right hand side and equating numerators yieldsx2 x+2=A(x2+4) + (Bx+C)xwhich can be solved (try it!) to obtainA=12,B=12,C= 1It follows that x2 x+2x3+4xdx= 12x+x 22(x2+4)dx=12ln|x|+ x2(x2+4) 1x2+4dx=12ln|x|+14ln(x2+4) 12tan 1x2+cWe had to be a little creative with the quadratic term in order to find an 4: Repeated Quadratic Factors (very hard!)

7 IfQ(x)contains a repeated factor(ax2+bx+c)mwhereax2+bx+cis irreducible andm 2, theneach such expression yields themtermsA1x+B1ax2+bx+c+A2x+B2(ax2+bx+c) 2+ +Amx+Bm(ax2+bx+c)mEach term may be integrated similarly to Case 3: part by inspection, part by completing the +bx+ccannot be factored (overR) into linear : These examples are often very involved. Master Cases 1 and 2 first!5( Partial ) ExampleTo integratex3+2x2+4(x2+2x+5)2(x 3)4(x 2)2we first seek a Partial fraction de-composition:x3+2x2+4(x2+2x+5)2(x 3)4(x 2)2=Ax+Bx2+2x+5+Cx+D(x2+2x+5)2+Ex 3+F(x 3)2+G(x 3)3+H(x 3)4+Ix 2+J(x 2)2 This is long and messy. The first two terms may be integrated by completing the square and substi-tutingu=x+1x2+2x+5= (x+1)2+4=u2+1 The integral of these terms will then be a combination of expressions such astan 1u2,ln(u2+1),(u2+1) 1If you re interested in the solution, ask a computer to help: the mathematician in you should becomfortable believing that it could be done!

8 RationalizingA clever substitution can sometimes convert an irrational expression into a rational one, to which thepartial Fractions method may be example, the substitutionu3=x 7 (dx=3u2du) gives 3 x 7x+1dx= 3u3u3+8du= 3 24(u+2)(u2 2u+4)du=3u+lnu2 2u+4(u+2)2 2 3 tan 1u 1 3+c( Partial Fractions in here)=3(x 7)1/3+ln(x 7)2/3 2(x 7)1/3+4((x 7)1/3+2)2 2 3 tan 1(x 7)1/3 1 3+cA similar approach (substitutingu= x 2) rationalizes the integral 1(x 2)(x 2+ x 2)dx= 2 duu2(u+1)Suggested problems1. Evaluate the integrals:(a) 8(x 2)(x+6)dx6(b) x(x 6)(x+2)2dx2. Evaluate the integrals:(a) 218 x2x(x2+5x+8)dx(b) 1y4+3y2+1dy3. Evaluate dxx2 1in two ways: using Partial Fractions and using a trigonometric your two up the integral of csc if you need to.

9 7