7524 x 2 x x 2 4 x 4 A x 2 B x 2 C x 4 yields A x 2 6 x 8 B x 2 2 x 8 C x 2 4 x

# 7524 x 2 x x 2 4 x 4 a x 2 b x 2 c x 4 yields a x 2 6

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7.5.24: x 2 + x ( x 2 4)( x + 4) = A x 2 + B x + 2 + C x + 4 yields A ( x 2 + 6 x + 8) + B ( x 2 + 2 x 8) + C ( x 2 4) = x 2 + x, so that A + B + C = 1, 6 A + 2 B = 1, and 8 A 8 B 4 C = 0. It follows that A = 1 4 , B = 1 4 , and C = 1. Therefore x 2 + x ( x 2 4)( x + 4) dx = 1 4 1 x 2 1 x + 2 + 4 x + 4 dx = 1 4 (ln | x 2 | − ln | x + 2 | + 4 ln | x + 4 | ) + C. 7.5.25: 1 x 3 + x = A x + Bx + C x 2 + 1 , so Ax 2 + A + Bx 2 + Cx = 1. Thus A + B = 0, C = 0, and A = 1. Therefore 1 x 3 + x dx = 1 x x x 2 + 1 dx = ln | x | − 1 2 ln( x 2 + 1) + C = 1 2 ln x 2 x 2 + 1 + C. 7.5.26: 6 x 3 18 x ( x 2 1)( x 2 4) = A x 2 + B x 1 + C x + 1 + D x + 2 leads to A ( x 3 + 2 x 2 x 2) + B ( x 3 + x 2 4 x 4) + C ( x 3 x 2 4 x + 4) + D ( x 3 2 x 2 x + 2) = 6 x 3 18 x. Thus A + B + C + D = 6 , 2 A + B C 2 D = 0 , A 4 B 4 C D = 18 , 2 A 4 B + 4 C + 2 D = 0 . It follows that A = 1, B = 2, C = 2, and D = 1. Therefore 6 x 3 18 x ( x 2 1)( x 2 4) dx = 1 x 2 + 2 x 1 + 2 x + 1 + 1 x + 2 dx = ln | x 2 | + 2 ln | x 1 | + 2 ln | x + 1 | + ln | x + 2 | + C. 691
7.7.27: x + 4 x 3 + 4 x = A x + Bx + C x 2 + 4 leads to Ax 2 + 4 A + Bx 2 + Cx = x + 4. So A + B = 0 , C = 1 , and 4 A = 4 . So A = 1 , B = 1 . Thus x + 4 x 3 + 4 x dx = 1 x x x 2 + 4 + 1 x 2 + 4 dx = ln | x | − 1 2 ln( x 2 + 4) + 1 2 arctan x 2 + C. 7.5.28: 4 x 4 + x + 1 x 5 + x 4 = A x + B x 2 + C x 3 + D x 4 + E x + 1 implies that A ( x 4 + x 3 ) + B ( x 3 + x 2 ) + C ( x 2 + x ) + D ( x + 1) + Ex 4 = 4 x 4 + x + 1 . Thus A + E = 4 , A + B = 0 , B + C = 0 , C + D = 1 , D = 1 . These equations are easily solve from the bottom up: D = 1, C = 0, B = 0, A = 0, and E = 4. Therefore 4 x 4 + x + 1 x 5 + x 4 dx = 1 x 4 + 4 x + 1 dx = 1 3 x 3 + 4 ln | x + 1 | + C. 7.5.29: x ( x + 1)( x 2 + 1) = A x + 1 + Bx + C x 2 + 1 yields Ax 2 + A + Bx 2 + Bx + Cx + C = x . Thus A + B = 0 , B + C = 1 , and A + C = 0 . It follows that A = 1 2 , B = 1 2 , and C = 1 2 . Therefore x ( x + 1)( x 2 + 1) dx = 1 2 1 x + 1 + x + 1 x 2 + 1 dx = 1 2 ln | x + 1 | + 1 4 ln( x 2 + 1) + 1 2 arctan x + C. 7.5.30: Rather than searching for unknown coeﬃcients, we note that x + 2 x 2 + 4 2 = x 2 + 4 x + 4 ( x 2 + 4) 2 = x 2 + 4 ( x 2 + 4) 2 + 4 x ( x 2 + 4) 2 = 1 x 2 + 4 + 4 x ( x 2 + 4) 2 . 692
Therefore x + 2 x 2 + 4 2 dx = 1 x 2 + 4 dx + 4 x ( x 2 + 4) 2 dx = 1 2 tan 1 x 2 2 x 2 + 1 + C. 7.5.31: x 2 10 2 x 4 + 9 x 2 + 4 = A x 2 + 4 + B 2 x 2 + 1 implies that 2 Ax 2 + A + Bx 2 + 4 B = x 2 10, and thus 2 A + B = 1 and A + 4 B = 10 , so that A = 2 and B = 3 . Therefore x 2 10 2 x 4 + 9 x 2 + 4 dx = 2 x 2 + 4 3 2 x 2 + 1 dx = arctan x 2 3 2 2 arctan x 2 + C. A substitution to integrate the second fraction is u = x 2. 7.5.32: x 2 x 4 1 = A x 1 + B x + 1 + Cx + D x 2 + 1 , so that A ( x 3 + x 2 + x + 1) + B ( x 3 x 2 + x 1) + C ( x 3 x ) + D ( x 2 1) = x 2 . Therefore A + B + C = 0 , A B + D = 1 , A + B C = 0 , A B D = 0 . It follows that A = 1 4 , B = 1 4 , C = 0, and D = 1 2 . Thus x 2 x 4 1 dx = 1 4 1 x 1 1 x + 1 + 2 x 2 + 1 dx = 1 4 ln | x 1 | − 1 4 ln | x + 1 | + 1 2 arctan x + C. 7.5.33: x 3 + x 2 + 2 x + 3 x 4 + 5 x 2 + 6 = Ax + B x 2 + 2 + Cx + D x 2 + 3 , and so Ax 3 + 3 Ax + Bx 2 + 3 B + Cx 3 + 2 Cx + Dx 2 + 2 D = x 3 + x 2 + 2 x + 3 . Therefore A + C = 1 , B + D = 1 , 3 A + 2 C = 2 , 3 B + 2 D = 3 . 693
It follows that A = 0, B = 1, C = 1, and D = 0. Hence x 3 + x 2 + 2 x + 3 x 4 + 5 x 2 + 6 dx = 1 x 2 + 2 + x x 2 + 3 dx = 2 2 arctan x 2 2 + 1 2 ln( x 2 + 3) + C.