fa07hw3

# fa07hw3 - 9 x-2 dx x-2 x-2 =x d x x2 x 2 A B Ax A B x 2B...

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Unformatted text preview: 9. x -2 dx +x -2 x -2 =x- d x. x2 + x - 2 A B Ax - A + B x + 2B x -2 = + = , then A+B = 1 and - A+2B = -2, If 2 x +x -2 x +2 x -1 x2 + x - 2 so that A = 4/3 and B = -1/3. Thus 1- x2 x2 dx 4 1 dx dx =x- + x2 + x - 2 3 x +2 3 x -1 1 4 = x - ln |x + 2| + ln |x - 1| + C. 3 3 x2 dx = x2 + x - 2 6. 1 B A + = 2 5-x 5-x 5+x (A + B) 5 + (A - B)x = 5 - x2 1 1 A + B = 5 A = B = . 2 5 A-B =0 1 1 1 1 dx + dx = 5 - x2 5-x 5+x 2 5 1 = - ln | 5 - x| + ln | 5 + x| + C 2 5 5+x 1 + C. = ln 2 5 5-x 4. x2 dx = x -4 = x +4+ 16 dx x -4 x2 + 4x + 16 ln |x - 4| + C. 2 41. For intersection of y = x4 9 and y = 1 we have + 4x 2 + 4 x 4 + 4x 2 + 4 = 9 x 4 + 4x 2 - 5 = 0 (x 2 + 5)(x 2 - 1) = 0, so the intersections are at x = 1. The required area is A=2 9 dx - 1 dx + 4x 2 + 4 0 1 dx = 18 - 2 Let x = 2 tan 2 2 0 (x + 2) d x = 2 sec2 x=1 2 sec2 d = 18 -2 4 sec4 x=0 x=1 9 cos2 d - 2 = 2 x=0 x=1 9 = + sin cos -2 2 2 x=0 1 9 2x -1 x -2 = tan + 2 2 2 2 x +2 0 x4 1 9 = tan-1 2 2 2 y y= 9 x 4 +4x 2 +4 1 - 1 units2 2 R y=1 -1 1 x Fig. 2-41 40. Area = 1 dx dx = 2 2x - x 1/2 1/2 1 - (x - 1)2 Let u = x - 1 du = d x 0 0 du -1 = = sin u 1 - u2 -1/2 -1/2 = sq. units. =0- - 6 6 1 0 33. - ln 2 e x 1 - e2x d x /2 Let e x = sin e x d x = cos d /2 /6 = /6 cos2 d = 1 = 2 3 - 3 4 1 ( + sin cos ) 2 3 = - . 6 8 20. = x2 x dx = - 2x + 3 u du + u2 + 2 u2 (x - 1) + 1 dx (x - 1)2 + 2 Let u = x - 1 du = d x du +2 1 1 u = ln(u 2 + 2) + tan-1 +C 2 2 2 1 x -1 1 = ln(x 2 - 2x + 3) + tan-1 2 2 2 + C. 9. = x3 dx 9 + x2 1 2 1 = u 3/2 - 9u 1/2 + C 3 1 = (9 + x 2 )3/2 - 9 9 + x 2 + C. 3 Let u = 9 + x 2 du = 2x d x (u - 9) du 1 (u 1/2 - 9u -1/2 ) du = 2 u 4. d x = cos d cos d = = csc d sin 1 - sin2 1 - 4x 2 1 - +C = ln | csc - cot | + C = ln 2x 2x 1 - 1 - 4x 2 + C1 . = ln x dx x 1 - 4x 2 Let x = 1 2 1 2 sin 31. In = (ln x)n d x U = (ln x)n dV = dx V =x dx dU = n(ln x)n-1 x n In = x(ln x) - n In-1 . I4 = x(ln x)4 - 4I3 = x(ln x)4 - 4 x(ln x)3 - 3I2 = x(ln x)4 - 4x(ln x)3 + 12 x(ln x)2 - 2I1 = x(ln x)4 - 4x(ln x)3 + 12x(ln x)2 - 24 x ln x - x + C = x (ln x)4 - 4(ln x)3 + 12(ln x)2 - 24 ln x + 24 + C. 21. ln(ln x) dx x = ln u du U = ln u du dU = u = u ln u - Let u = ln x dx du = x d V = du V =u du = u ln u - u + C = (ln x)(ln(ln x)) - ln x + C. e 20. I = 1 sin(ln x) d x U = sin(ln x) cos(ln x) dU = dx x e e dV = dx V =x = x sin(ln x) - 1 1 cos(ln x) d x dV = dx V =x e 1 U = cos(ln x) sin(ln x) dx dU = - x = e sin(1) - x cos(ln x) + I 1 Thus, I = [e sin(1) - e cos(1) + 1]. 2 ...
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