🔗 Activity 5.6.1. 🔗 🔗Consider .∫x2+x+1x3+xdx. Which substitution would you choose to evaluate this integral? 🔗u=x3 🔗u=x3+x 🔗u=x2+x+1 🔗Substitution is not effective
🔗 Activity 5.6.2. 🔗 🔗Using the method of substitution, which of these is equal to ?∫5x+7dx? 🔗5ln|x+7|+C 🔗57ln|x+7|+C 🔗5ln|x|+5ln|7|+C 🔗57ln|x|+C
🔗 Observation 5.6.3. 🔗 🔗To avoid repetitive substitution, the following integral formulas will be useful. ∫1x+bdx=ln|x+b|+C ∫1(x+b)2dx=−1x+b+C ∫1x2+b2dx=1barctan(xb)+C
🔗 Activity 5.6.4. 🔗 🔗Which of the following is equal to ?1x+1x2+1? 🔗2xx2+x+1 🔗x3+xx2+x+1 🔗2xx3+x 🔗x2+x+1x3+x
🔗 Activity 5.6.5. 🔗 🔗Based on the previous activities, which of these is equal to ?∫x2+x+1x3+xdx? 🔗ln|x|+arctan(x)+C 🔗ln|x2+x+1|+C 🔗ln|x3+x|+C 🔗arctan(x3+x)+C
🔗 Activity 5.6.6. 🔗 🔗Suppose we know .10x−11x2+x−2=7x−1+3x+2. 🔗Which of these is equal to ?∫10x−11x2+x−2? 🔗7ln|x−1|+3arctan(x+2)+C 🔗7ln|x−1|+3ln|x+2|+C 🔗7arctan(x−1)+3arctan(x+2)+C 🔗7arctan(x−1)+3ln|x+2|+C
🔗 Observation 5.6.7. 🔗 🔗To find integrals like ∫x2+x+1x3+xdx and ,∫10x−11x2+x−2, we'd like to decompose the fractions into simpler partial fractions that may be integrated with these formulas ∫1x+bdx=ln|x+b|+C ∫1(x+b)2dx=−1x+b+C ∫1x2+b2dx=1barctan(xb)+C
🔗 Fact 5.6.8. Partial Fraction Decomposition. 🔗Let p(x)q(x) be a rational function, where the degree of p is less than the degree of .q. 🔗 🔗 Linear Terms: Let (x−a)n divide ,q(x), Then the decomposition of p(x)q(x) will contain the terms .A1(x−a)+A2(x−a)2+⋯+An(x−a)n. 🔗 Quadratic Terms: Let (x2+bx+c)n divide ,q(x), where x2+bx+c is irreducable. Then the decomposition of p(x)q(x) will contain the terms .B1x+C1x2+bx+c+B2x+C2(x2+bx+c)2+⋯+Bnx+Cn(x2+bx+c)n.
🔗 Example 5.6.9. 🔗 🔗Following is an example of a rather involved partial fraction decomposition. 7x6−4x5+41x4−20x3+24x2+11x+16x(x−1)2(x2+4)2=Ax+Bx−1+C(x−1)2+Dx+Ex2+4+Fx+G(x2+4)2 🔗Using some algebra, it's possible to find values for A through G to determine .7x6−4x5+41x4−20x3+24x2+11x+16x(x−1)2(x2+4)2=1x+2x−1+3(x−1)2+4x+5x2+4+6x+7(x2+4)2.
🔗 Activity 5.6.10. 🔗 🔗Which of the following is the form of the partial fraction decomposition of ?x3−7x2−7x+15x3(x+5)? 🔗Ax+Bx+5 🔗Ax3+Bx+5 🔗Ax+Bx2+Cx3+Dx+5 🔗Ax+Bx2+Cx3+Dx+Ex+5
🔗 Activity 5.6.11. 🔗 🔗Which of the following is the form of the partial fraction decomposition of ?x2+1(x−3)2(x2+4)2? 🔗Ax−3+B(x−3)2+Cx2+4+D(x2+4)2 🔗Ax−3+B(x−3)2+Cx+D(x2+4)2 🔗Ax−3+B(x−3)2+Cx2+4+Dx+E(x2+4)2 🔗Ax−3+B(x−3)2+Cx+Dx2+4+Ex+F(x2+4)2
🔗 Activity 5.6.12. 🔗 🔗Consider that the partial decomposition of x2+5x+3(x+1)2x is x2+5x+3(x+1)2x=Ax+1+B(x+1)2+Cx. 🔗What equality do we obtain if we multiply both sides of the above equation by ?(x+1)2x? 🔗x2+5x+3=Ax(x+1)+Bx+C(x+1)2 🔗x2+5x+3=A(x+1)+B(x+1)2+Cx 🔗x2+5x+3=Ax(x+1)+Bx+C(x+1) 🔗x2+5x+3=Ax(x+1)+Bx2+C(x+1)2
🔗 Activity 5.6.13. 🔗Use your choice in Activity 5.6.12 (which must hold for any x value) to answer the following. 🔗(a) 🔗By substituting x=0 into the equation, we may find: 🔗A=1 🔗B=−2 🔗C=3 🔗(b) 🔗By substituting x=−1 into the equation, we may find: 🔗A=−4 🔗B=1 🔗C=5
🔗 Activity 5.6.14. 🔗 🔗Using the results of Activity 5.6.13, show how to rewrite our choice from Activity 5.6.12 .?x2+?x=Ax2+Ax. 🔗What value of A satisfies this equation? 🔗−2 🔗3 🔗4 🔗−5
🔗 Activity 5.6.15. 🔗By using the form of the decomposition x2+5x+3(x+1)2x=Ax+1+B(x+1)2+Cx and the coefficients found in Activity 5.6.13 and Activity 5.6.14, evaluate .∫x2+5x+3(x+1)2xdx.
🔗 Activity 5.6.16. 🔗Given that x3−7x2−7x+15x3(x+5)=Ax+Bx2+Cx3+Dx+5 do the following to find ,A,B,C, and .D. 🔗(a) 🔗 🔗Eliminate the fractions to obtain .x3−7x2−7x+15=A(?)(?)+B(?)(?)+C(?)+D(?). 🔗(b) 🔗Plug in an x value that lets you find the value of .C. 🔗(c) 🔗Plug in an x value that lets you find the value of .D. 🔗(d) 🔗Use other algebra techniques to find the values of A and .B.
🔗 Activity 5.6.18. 🔗 🔗Consider the rational expression 2x3+2x+4x4+2x3+4x2. Which of the following is the partial fraction decomposition of this rational expression? 🔗1x+1x2+2x−1x2+2x+4 🔗2x+0x2+−1x2+2x+4 🔗0x+1x2+−1x2+2x+4 🔗0x+1x2+2x−1x2+2x+4