Testing the factors of 27, we nd that 3 is a root of D(x). Dividing D(x) by x 3 yields in .NET

Making QR Code JIS X 0510 in .NET Testing the factors of 27, we nd that 3 is a root of D(x). Dividing D(x) by x 3 yields

Testing the factors of 27, we nd that 3 is a root of D(x). Dividing D(x) by x 3 yields
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and so the partial fractions representation is (Case 2) A2 A3 x2 A1 + + = 2 (x + 3) 2 x 3 (x 3) x+3 (x 3) Multiply by (x 3)2 (x + 3), x 2 = A1 (x 3)(x + 3) + A2 (x + 3) + A3 (x 3)2 3 9 = 0 + 6A2 + 0 or A2 = 2 9 = 0 + 0 + A3 ( 6)2 or A3 = 1 4
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Let x = 3, Let x = 3, Compare coef cients of x 2 ,
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1 = A1 + A3
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A1 = 1 A3 =
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THE METHOD OF PARTIAL FRACTIONS
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[CHAP. 40
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Thus,
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3 1 1 1 x2 3 1 + + = 3 3x 2 9x + 27 4 x 3 2 (x 3)2 4 x + 3 x x 2 dx x 3 3x 2 9x + 27 = 3 3 1 1 ln |x 3| + ln |x + 3| + C 4 2x 3 4
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40.3 Find
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x+1 dx. x(x 2 + 2)
x+1 A A x + A3 = 1 + 22 x x(x 2 + 2) x +2
This is Case 3,
Multiply by x(x 2 + 2), x + 1 = A1 (x 2 + 2) + x(A2 x + A3 ) 1 2
Let x = 0, Compare coef cients of x 2 ,
1 = 2A1 + 0
A1 =
0 = A1 + A2 Compare coef cients of x,
A2 = A1 =
Thus, and
1 = A3 ( 1 )x + 1 x+1 1 1 2 + = 2 x x(x 2 + 2) x2 + 2 x+1 1 1 x dx 1 dx dx dx = + 2 + 2) 2+2 2+2 2 x 2 x(x x x
Because the quadratic factor x 2 + 2 is a complete square, we can perform the integrations on the right without a change of variable, x+1 1 1 x 1 dx = ln |x| ln (x 2 + 2) + tan 1 + C 2 4 x(x 2 + 2) 2 2
40.4 Evaluate
1 dx. 1 sin x + cos x
Observe that the integrand is a rational function of sin x and cos x. Any rational function of the six trigonometric functions reduces to a function of this type, and the method we shall use to solve this particular problem will work for any such function. Make the change of variable z = tan (x/2); that is, x = 2 tan 1 z. Then, dx = and, by Theorem 26.8, sin x = 2 sin x tan (x/2) x cos = 2 2 2 2 sec (x/2) tan (x/2) 2z =2 = 1 + z2 1 + tan2 (x/2) tan2 (x/2) x =1 2 2 2 sec (x/2) 2 dz 1 + z2
cos x = 1 2 sin2 =1 2
2z2 tan2 (x/2) 1 z2 =1 = 2 (x/2) 2 1+z 1 + z2 1 + tan
CHAP. 40]
THE METHOD OF PARTIAL FRACTIONS
When these substitutions are made, the resulting integrand will be a rational function of z (because compositions and products of rational functions are rational functions). The method of partial fractions can then be applied, (1 sin x + cos x) 1 dx = = = 1 2z 1 z2 1 2 + dz 1 + z2 1 + z2 1 + z2
(1 + z2 ) 2z + (1 z2 ) 1 + z2
2 dz 1 + z2
1 + z2 2 2 2z 1 2 dz = dz 2 2 2 2z 1 + z2 1+z 1+z 1 dz = ln |1 z| + C = 1 z x +C = ln 1 tan 2
40.5 Find
x dx . (x + 1)(x 2 + 2x + 2)2
x (x + 1)(x 2 + 2x + 2)2 = A x + A3 A1 A x + A5 + 22 + 2 4 x + 1 x + 2x + 2 (x + 2x + 2)2
This is Case 4 for D(x), and so
Multiply by (x + 1)(x 2 + 2x + 2)2 , x = A1 (x 2 + 2x + 2)2 + (A2 x + A3 )(x + 1)(x 2 + 2x + 2) + (A4 x + A5 )(x + 1) or, partially expanding the right-hand side, x = A1 (x 4 + 4x 3 + 8x 2 + 8x + 4) + (A2 x + A3 )(x 3 + 3x 2 + 4x + 2) + (A4 x + A5 )(x + 1) In (1), let x = 1, Compare coef cients of x 4 , 0 = A1 + A2 Compare coef cients of x 3 , 0 = 4A1 + 3A2 + A3 Compare coef cients of x 2 , 0 = 8A1 + 4A2 + 3A3 + A4 Compare coef cients of x 0 , 0 = 4A1 + 2A3 + A5 Therefore, x dx = 1 (x + 1)(x 2 + 2x + 2)2 dx + x+1 (x + 2) dx (x + 1) dx + 2 + 2x + 2 2 + 2x + 2)2 x (x u du du u du + + u2 + 1 (u2 + 1)2 (u2 + 1)2 [by Quick Formulas II and I] 1 1 1 + = ln |x + 1| + ln (u2 + 1) 2 2 u2 + 1 sin 2 1 1 1 +C = ln |x + 1| + ln (x 2 + 2x + 2) + + 2 2 x 2 + 2x + 2 2 4 Now = tan 1 u = tan 1 (x + 1) cos2 d [Case 4: let u = tan ] or A5 = 4A1 2A3 = 2 or A4 = 8A1 4A2 3A3 = 1 or A3 = 4A1 3A2 = 1 or A2 = A1 = 1 1 = A1 (1) = A1 (1)
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