7534 x 2 4 x 2 1 2 x 2 2 Ax B x 2 2 Cx D x 2 1 Ex F x 2 1 2 yields A x 5 2 x 3

# 7534 x 2 4 x 2 1 2 x 2 2 ax b x 2 2 cx d x 2 1 ex f x

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7.5.34: x 2 + 4 ( x 2 + 1) 2 ( x 2 + 2) = Ax + B x 2 + 2 + Cx + D x 2 + 1 + Ex + F ( x 2 + 1) 2 yields A ( x 5 + 2 x 3 + x ) + B ( x 4 + 2 x 2 + 1) + C ( x 5 + 3 x 3 + 2 x ) + D ( x 4 + 3 x 2 + 2) + E ( x 3 + 2 x ) + F ( x 2 + 2) = x 2 + 4 . Therefore A + C = 0 , B + D = 0 , 2 A + 3 C + E = 0 , 2 B + 3 D + F = 1 , A + 2 C + 2 E = 0 , B + 2 D + 2 F = 4 . Note how the equations involving A , C , and E “separate” from those involving B , D , and F . This makes it easy to solve them for A = 0, B = 2, C = 0, D = 2, E = 0, and F = 3. Therefore x 2 + 4 ( x 2 + 1) 2 ( x 2 + 2) dx = 2 x 2 + 2 2 x 2 + 1 + 3 ( x 2 + 1) 2 dx = 2 arctan x 2 2 1 2 arctan x + 3 x 2( x 2 + 1) + C. (Part of the solution of Problem 30 was used to integrate the third fraction.) 7.5.35: Expand the denominator to x 4 2 x 3 + 2 x 2 2 x + 1 and then divide it into the numerator to find that x 4 + 3 x 2 4 x + 5 ( x 2 + 1)( x 1) 2 = 1 + 2 x 3 + x 2 2 x + 4 ( x 2 + 1)( x 1) 2 . Then 2 x 3 + x 2 2 x + 4 ( x 2 + 1)( x 1) 2 = A x 1 + B ( x 1) 2 + Cx + D x 2 + 1 leads to A ( x 3 x 2 + x 1) + B ( x 2 + 1) + C ( x 3 2 x 2 + x ) + D ( x 2 2 x + 1) = 2 x 3 + x 2 2 x + 4 . Therefore A + C = 2 , A + B 2 C + D = 1 , A + C 2 D = 2 , A + B + D = 4; 694
the solution is A = 1 2 , B = 5 2 , C = 3 2 , D = 2. Therefore x 4 + 3 x 2 4 x + 5 ( x 2 + 1)( x 1) 2 dx = 1 + 1 2 x 1 + 5 2 ( x 1) 2 + 3 2 x x 2 + 1 + 2 x 2 + 1 dx = x + 1 2 ln | x 1 | − 5 2( x 1) + 3 4 ln( x 2 + 1) + 2 arctan x + C. 7.5.36: 2 x 3 + 5 x 2 x + 3 ( x 2 + x 2) 2 = A x 1 + B ( x 1) 2 + C x + 2 + D ( x + 2) 2 yields A ( x 1)( x 2 + 4 x + 4) + B ( x 2 + 4 x + 4) + C ( x + 2)( x 2 2 x + 1) + D ( x 2 2 x + 1) = 2 x 3 + 5 x 2 x + 3; A ( x 3 + 3 x 2 4) + B ( x 2 + 4 x + 4) + C ( x 3 3 x + 2) + D ( x 2 2 x + 1) = 2 x 3 + 5 x 2 x + 3 . Thus A + C = 2 , 3 A + B + D = 5 , 4 B 3 C 2 D = 1 , 4 A + 4 B + 2 C + D = 3 . It follows that A = B = C = D = 1, and therefore 2 x 3 + 5 x 2 x + 3 ( x 2 + x 2) 2 dx = 1 x 1 + 1 ( x 1) 2 + x x + 2 + 1 ( x + 2) 2 dx = ln | x 1 | − 1 x 1 + ln | x + 2 | − 1 x + 2 + C. 7.5.37: Let x = e 2 t ; then dx = 2 e 2 t dt . Thus e 4 t ( e 2 t 1) 3 dt = 1 2 x ( x 1) 3 dx. Then x ( x 1) 3 = A x 1 + B ( x 1) 2 + C ( x 1) 3 leads to A ( x 2 2 x + 1) + B ( x 1) + C = x , and so A = 0 , 2 A + B = 1 , and A B + C = 0 . Therefore B = C = 1. Hence x ( x 1) 3 dx = 1 ( x 1) 2 + 1 ( x 1) 3 dx = 1 x 1 1 2( x 1) 2 + C. Finally, e 4 t ( e 2 t 1) 3 dt = 1 2( x 1) 1 4( x 1) 2 + C = 1 2( e 2 t 1) 1 4( e 2 t 1) 2 + C. 695
7.5.38: Let u = sin θ ; then du = cos θ dθ . So I = cos θ sin 2 θ sin θ 6 = 1 u 2 u 6 du = 1 5 1 u 3 1 u + 2 du = 1 5 (ln | u 3 | − ln | u + 2 | ) + C = 1 5 (ln | − 3 + sin θ | − ln | 2 + sin θ | ) + C. 7.5.39: Let u = ln t ; then du = 1 t dt . Therefore J = 1 + ln t t (3 + 2 ln t ) 2 dt = u + 1 (2 u + 3) 2 du = 1 2 1 2 u + 3 1 (2 u + 3) 2 du = 1 4 ln | 2 u + 3 | + 1 4 · 1 2 u + 3 + C = 1 4 ln | 3 + 2 ln t | + 1 4(3 + 2 ln t ) + C. 7.5.40: Let u = tan t . Then du = sec 2 t dt . Hence K = sec 2 t tan 3 t + tan 2 t dt = 1 u 3 + u 2 du = 1 u + 1 u 2 + 1 u + 1 du = ln | u | − 1 u + ln | 1 + u | + C = ln 1 + tan t tan t cot t + C = ln cos t + sin t sin t cot t + C = ln | sin t + cos t | − ln | sin t | − cot t + C. 7.5.41: x 9 x 2 3 x = 3 x 2 x 3 . So 2 1 x 9 x 2 3 x dx = 3 ln | x | − 2 ln | x 3 | 2 1 = 3 ln 2 ( 2 ln 2) = 5 ln 2 3 . 4657359028 .