sol231121

23-November-2021 Homework:

Part I: Compute the following indefinite integrals (Hint: one can use partial fraction decomposition for rational functions). (You must give some necessary steps, and cannot directly write down the answer.)

1. %5E2(x%2B4)%5E2%7D%20dx#crop=0&crop=0&crop=1&crop=1&id=EkO5U&originHeight=52&originWidth=179&originalType=binary&ratio=1&rotation=0&showTitle=false&status=done&style=none&title=)
2. (x%2B2)%5E2(x%2B3)%5E3%7D%20dx#crop=0&crop=0&crop=1&crop=1&id=qZmgI&originHeight=48&originWidth=236&originalType=binary&ratio=1&rotation=0&showTitle=false&status=done&style=none&title=)
3. %5E3%7D%20dx#crop=0&crop=0&crop=1&crop=1&id=Um8wg&originHeight=48&originWidth=122&originalType=binary&ratio=1&rotation=0&showTitle=false&status=done&style=none&title=)

Solution to Part I.

1. Let us firstly determine the partial fraction decomposition for the integrand. Suppose
   %5E2(x%2B4)%5E2%7D%20%3D%20%5Cfrac%7BA%7D%7Bx%2B2%7D%2B%5Cfrac%7BB%7D%7B(x%2B2)%5E2%7D%2B%5Cfrac%7BC%7D%7Bx%2B4%7D%2B%5Cfrac%7BD%7D%7B(x%2B4)%5E2%7D#card=math&code=%5Cfrac%7Bx%5E2%7D%7B%28x%2B2%29%5E2%28x%2B4%29%5E2%7D%20%3D%20%5Cfrac%7BA%7D%7Bx%2B2%7D%2B%5Cfrac%7BB%7D%7B%28x%2B2%29%5E2%7D%2B%5Cfrac%7BC%7D%7Bx%2B4%7D%2B%5Cfrac%7BD%7D%7B%28x%2B4%29%5E2%7D&id=V68yL). Clear the denominators we get (x%2B4)%5E2%20%2B%20B(x%2B4)%5E2%2BC(x%2B2)%5E2(x%2B4)%2BD(x%2B2)%5E2.%0A#card=math&code=x%5E2%3DA%28x%2B2%29%28x%2B4%29%5E2%20%2B%20B%28x%2B4%29%5E2%2BC%28x%2B2%29%5E2%28x%2B4%29%2BD%28x%2B2%29%5E2.%0A&id=w3DGO) (1)
   Let , the LHS , the RHS . So .
   Similarly, letting , we deduce that .
   To determine and , we take derivative on both hand-sides of the eq. (1), and obtain %5E2%20%2B%202A(x%2B2)(x%2B4)%2B2B(x%2B4)%2BC(x%2B2)%5E2%2B2C(x%2B2)(x%2B4)%2B2D(x%2B2).%0A#card=math&code=2x%20%3D%20A%28x%2B4%29%5E2%20%2B%202A%28x%2B2%29%28x%2B4%29%2B2B%28x%2B4%29%2BC%28x%2B2%29%5E2%2B2C%28x%2B2%29%28x%2B4%29%2B2D%28x%2B2%29.%0A&id=jVzgO)
   Now letting , the LHS , and the RHS . So .
   Similarly, letting , we deduce that .
   So %5E2(x%2B4)%5E2%7D%20%3D%20%5Cfrac%7B-2%7D%7Bx%2B2%7D%2B%5Cfrac%7B1%7D%7B(x%2B2)%5E2%7D%2B%5Cfrac%7B2%7D%7Bx%2B4%7D%2B%5Cfrac%7B4%7D%7B(x%2B4)%5E2%7D.%0A#card=math&code=%5Cfrac%7Bx%5E2%7D%7B%28x%2B2%29%5E2%28x%2B4%29%5E2%7D%20%3D%20%5Cfrac%7B-2%7D%7Bx%2B2%7D%2B%5Cfrac%7B1%7D%7B%28x%2B2%29%5E2%7D%2B%5Cfrac%7B2%7D%7Bx%2B4%7D%2B%5Cfrac%7B4%7D%7B%28x%2B4%29%5E2%7D.%0A&id=PRJO6)
   Then, the anti-derivative is
   .

2. Suppose the partial fraction decomposition is %5E2%7D%2B%5Cfrac%7BD%7D%7Bx%2B3%7D%2B%5Cfrac%7BE%7D%7B(x%2B3)%5E2%7D%2B%5Cfrac%7BF%7D%7B(x%2B3)%5E3%7D%3D%5Cfrac%7B1%7D%7B(x%2B1)(x%2B2)%5E2(x%2B3)%5E3%7D.%20%0A#card=math&code=%5Cfrac%7BA%7D%7Bx%2B1%7D%2B%5Cfrac%7BB%7D%7Bx%2B2%7D%2B%5Cfrac%7BC%7D%7B%28x%2B2%29%5E2%7D%2B%5Cfrac%7BD%7D%7Bx%2B3%7D%2B%5Cfrac%7BE%7D%7B%28x%2B3%29%5E2%7D%2B%5Cfrac%7BF%7D%7B%28x%2B3%29%5E3%7D%3D%5Cfrac%7B1%7D%7B%28x%2B1%29%28x%2B2%29%5E2%28x%2B3%29%5E3%7D.%20%0A&id=WvDMi) (2)
   To determine the coefficients, we proceed as follows.
   Multiplying on both hand-sides of (2), we have
   #card=math&code=A%20%2B%20%28x%2B1%29&id=r9APD)(the rest terms) = %5E2(x%2B3)%5E3%7D#card=math&code=%5Cfrac%7B1%7D%7B%28x%2B2%29%5E2%28x%2B3%29%5E3%7D&id=BXwnB),
   where the rest terms have will not tend to when . By letting on both sides, we get that .

Multiplying %5E2#card=math&code=%28x%2B2%29%5E2&id=NRq4V) on both hand-sides of (2), we have
%2BC%2B%20(x%2B2)%5E2#card=math&code=B%28x%2B2%29%2BC%2B%20%28x%2B2%29%5E2&id=zgKN3)(the rest terms)= (x%2B3)%5E3%7D#card=math&code=%5Cfrac%7B1%7D%7B%28x%2B1%29%28x%2B3%29%5E3%7D&id=PrrCD),
where the rest terms will not tend to as . By letting , we get that . Now taking derivative with respect to on both sides from the last equation, we have
#card=math&code=B%2B2%28x%2B2%29&id=DxKIv)(the rest terms) = %5E%7B-2%7D(x%2B3)%5E%7B-3%7D-3(x%2B1)(x%2B3)%5E%7B-4%7D#card=math&code=-%28x%2B1%29%5E%7B-2%7D%28x%2B3%29%5E%7B-3%7D-3%28x%2B1%29%28x%2B3%29%5E%7B-4%7D&id=PCWJc).

By letting we get that .

Using same method, we multiplying %5E3#card=math&code=%28x%2B3%29%5E3&id=uUueu) on both hand-sides of (2), we deduce that .
So the anti-derivative should be

1. (a) It is not easy to compute %5E3%7D#card=math&code=I_3%3D%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28x%5E2%2B9%29%5E3%7D&id=Z7cLT) via partial fraction decomposition. Instead, one can use integration by parts: %5E3%7D%20%3D%20%5Cfrac%7Bx%7D%7B(x%5E2%2B9)%5E3%7D%20%2B6%5Cint%20%5Cfrac%7Bx%5E2%7D%7B(x%5E2%2B9)%5E4%7D%20dx#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28x%5E2%2B9%29%5E3%7D%20%3D%20%5Cfrac%7Bx%7D%7B%28x%5E2%2B9%29%5E3%7D%20%2B6%5Cint%20%5Cfrac%7Bx%5E2%7D%7B%28x%5E2%2B9%29%5E4%7D%20dx&id=oPV4z) . This suggests that one should instead consider the integral %5E2%7Dd%20x#card=math&code=I_2%20%3D%5Cint%20%5Cfrac%7B1%7D%7B%28x%5E2%2B9%29%5E2%7Dd%20x&id=XLiyu) rather. Applying the integration by parts technique to , we have
   %5E2%7D%20%2B4%5Cint%20%5Cfrac%7Bx%5E2%7D%7B(x%5E2%2B9)%5E3%7D%20d%20x#card=math&code=I_2%20%3D%20%5Cfrac%7Bx%7D%7B%28x%5E2%2B9%29%5E2%7D%20%2B4%5Cint%20%5Cfrac%7Bx%5E2%7D%7B%28x%5E2%2B9%29%5E3%7D%20d%20x&id=B5U1I).
   We write the second summand as %5E3%7D%20dx%20%3D%204%20%5Cint%20%5Cfrac%7Bx%5E2%2B9%7D%7B(x%5E2%2B9)%5E3%7D%20dx%20-%2036%5Cint%20%5Cfrac%7B1%7D%7B(x%5E2%2B9)%5E3%7D%20dx#card=math&code=4%20%5Cint%20%5Cfrac%7Bx%5E2%7D%7B%28x%5E2%2B9%29%5E3%7D%20dx%20%3D%204%20%5Cint%20%5Cfrac%7Bx%5E2%2B9%7D%7B%28x%5E2%2B9%29%5E3%7D%20dx%20-%2036%5Cint%20%5Cfrac%7B1%7D%7B%28x%5E2%2B9%29%5E3%7D%20dx&id=ZGvbw). Thus, %5E2%7D#card=math&code=I_3%20%3D%20%5Cfrac%7BI_2%7D%7B12%7D%2B%5Cfrac%7Bx%7D%7B36%28x%5E2%2B9%29%5E2%7D&id=FQh0C).
   The same idea gives that %7D%2B%20I_1)%2C%20%5C%3B%20I_1%20%3D%20%5Cint%20%5Cfrac%7Bd%20x%7D%7Bx%5E2%2B9%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Carctan%20%5Cfrac%7Bx%7D%7B3%7D%2BC#card=math&code=I_2%20%3D%20%5Cfrac%7B1%7D%7B18%7D%28%5Cfrac%7Bx%7D%7B%28x%5E2%2B9%29%7D%2B%20I_1%29%2C%20%5C%3B%20I_1%20%3D%20%5Cint%20%5Cfrac%7Bd%20x%7D%7Bx%5E2%2B9%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Carctan%20%5Cfrac%7Bx%7D%7B3%7D%2BC&id=tUkN5).
   Therefore, we have %5E2%7D%2B%20%5Cfrac%7B1%7D%7B216%7D(%5Cfrac%7Bx%7D%7Bx%5E2%2B9%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%5Carctan%20%5Cfrac%7Bx%7D%7B3%7D)%2BC.%20%0A#card=math&code=I_3%20%3D%5Cfrac%7Bx%7D%7B36%28x%5E2%2B9%29%5E2%7D%2B%20%5Cfrac%7B1%7D%7B216%7D%28%5Cfrac%7Bx%7D%7Bx%5E2%2B9%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%5Carctan%20%5Cfrac%7Bx%7D%7B3%7D%29%2BC.%20%0A&id=MZYm6)

(b) We give one more methods to solve this problem. Set . Then

%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%20%5Cint%20%5Cfrac%7Bd%20(x%2F3)%7D%7B%5B1%2B(x%2F3)%5E2%5D%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%20%5Cint%20%5Cfrac%7Bd%20(%5Ctan%20u)%7D%7B(1%2B%5Ctan%5E2%20u)%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%5Cint%20%5Ccos%5E4%20u%5C%2C%20d%20u.%0A#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28x%5E2%2B9%29%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%20%5Cint%20%5Cfrac%7Bd%20%28x%2F3%29%7D%7B%5B1%2B%28x%2F3%29%5E2%5D%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%20%5Cint%20%5Cfrac%7Bd%20%28%5Ctan%20u%29%7D%7B%281%2B%5Ctan%5E2%20u%29%5E3%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E5%7D%5Cint%20%5Ccos%5E4%20u%5C%2C%20d%20u.%0A&id=nIzzr)

Using the identity twice, we get %2BC#card=math&code=%5Cfrac%7B1%7D%7B3%5E5%7D%5Cint%20%5Ccos%5E4%20u%5C%2C%20d%20u%20%3D%20%5Cfrac%7B1%7D%7B6%5E5%7D%5Cbig%28%5Csin%204u%20%2B%208%20%5Csin%202u%20%2B%2012%20u%20%5Cbig%29%2BC&id=f7Zp1).

Since , we can assume that . And hence .
So %7D%7B(9%2Bx%5E2)%5E2%7D#card=math&code=%5Csin%204%20u%20%3D%202%20%5Csin%202u%20%5Ccos%202u%20%3D%20%5Cfrac%7B12x%289-x%5E2%29%7D%7B%289%2Bx%5E2%29%5E2%7D&id=QKvVO). So we finally obtain

%5E3%7D%20%26%3D%5Cfrac%7B1%7D%7B6%5E5%7D%5Cbig(%5Csin%204u%20%2B%208%20%5Csin%202u%20%2B%2012%20u%20%5Cbig)%2BC%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B6%5E5%7D%5Cbigg(%5Cfrac%7B12x(9-x%5E2)%7D%7B(9%2Bx%5E2)%5E2%7D%20%2B%20%5Cfrac%7B48x%7D%7Bx%5E2%2B9%7D%2B%2012%20%5Carctan(%5Cfrac%7Bx%7D%7B3%7D)%5Cbigg)%2BC.%20%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28x%5E2%2B9%29%5E3%7D%20%26%3D%5Cfrac%7B1%7D%7B6%5E5%7D%5Cbig%28%5Csin%204u%20%2B%208%20%5Csin%202u%20%2B%2012%20u%20%5Cbig%29%2BC%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B6%5E5%7D%5Cbigg%28%5Cfrac%7B12x%289-x%5E2%29%7D%7B%289%2Bx%5E2%29%5E2%7D%20%2B%20%5Cfrac%7B48x%7D%7Bx%5E2%2B9%7D%2B%2012%20%5Carctan%28%5Cfrac%7Bx%7D%7B3%7D%29%5Cbigg%29%2BC.%20%0A%5Cend%7Baligned%7D%0A&id=KX6Pz)

Part II: Openstax Calculus textbook (volume 1):

Page 613, EXERCISES 5.7, Problems No. 397, 398, 399, 400, 423, 424, 425, 426.

Solution to Part II.

Prob. 397. %7D%7B%5Csqrt%7B1-(x%2F3)%5E2%7D%7D%20%3D%20%5Carcsin(%5Cfrac%7Bx%7D%7B3%7D)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%5Csqrt%7B9-x%5E2%7D%7D%20%3D%20%5Cint%20%5Cfrac%7Bd%20%28x%2F3%29%7D%7B%5Csqrt%7B1-%28x%2F3%29%5E2%7D%7D%20%3D%20%5Carcsin%28%5Cfrac%7Bx%7D%7B3%7D%29%2BC&id=E0lQQ).

Prob. 398. %5E2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint%20%5Cfrac%7Bd%20(4x)%7D%7B%5Csqrt%7B1-(4x)%5E2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%5Carcsin(4x)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%5Csqrt%7B1-%284x%29%5E2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint%20%5Cfrac%7Bd%20%284x%29%7D%7B%5Csqrt%7B1-%284x%29%5E2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%5Carcsin%284x%29%2BC&id=wDQol).

Prob. 399. %7D%7B1%2B(x%2F3)%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Carctan(%5Cfrac%7Bx%7D%7B3%7D)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7Bx%5E2%2B9%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cint%20%5Cfrac%7Bd%20%28x%2F3%29%7D%7B1%2B%28x%2F3%29%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Carctan%28%5Cfrac%7Bx%7D%7B3%7D%29%2BC&id=qE10z).

Prob. 400. .

Prob. 423. %7D%7B%5Csqrt%7B1-e%5E%7B2t%7D%7D%7D%20%3D%20%5Carcsin(e%5Et)%2BC#card=math&code=%5Cint%20%5Cfrac%7Be%5Et%7D%7B%5Csqrt%7B1-e%5E%7B2t%7D%7D%7D%20dt%20%3D%20%5Cint%20%5Cfrac%7B%20d%28e%5Et%29%7D%7B%5Csqrt%7B1-e%5E%7B2t%7D%7D%7D%20%3D%20%5Carcsin%28e%5Et%29%2BC&id=pcRjy).

Prob. 424. %2BC#card=math&code=%5Cint%20%5Cfrac%7Be%5Et%7D%7B1%2Be%5E%7B2t%7D%7Ddt%20%3D%20%5Carctan%28e%5Et%29%2BC&id=nNiZS).

Prob. 425. %7D%7B%5Csqrt%7B1-%5Cln%5E2%20t%7D%7D%3D%5Carcsin(%5Cln%20t)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20t%7D%7Bt%20%5Csqrt%7B1-%5Cln%5E2%20t%7D%7D%20%20%3D%20%5Cint%20%5Cfrac%7B%20d%20%28%5Cln%20t%29%7D%7B%5Csqrt%7B1-%5Cln%5E2%20t%7D%7D%3D%5Carcsin%28%5Cln%20t%29%2BC&id=aI8ek).

Prob. 426. %7D%20%3D%20%5Cint%20%5Cfrac%7Bd(%5Cln%20t)%7D%7B1%2B%5Cln%5E2%20t%7D%20%3D%20%5Carctan(%5Cln%20t)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20t%7D%7Bt%281%2B%5Cln%5E2%20t%29%7D%20%3D%20%5Cint%20%5Cfrac%7Bd%28%5Cln%20t%29%7D%7B1%2B%5Cln%5E2%20t%7D%20%3D%20%5Carctan%28%5Cln%20t%29%2BC&id=RkvYc).