By Riley, Hobson.

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A) Evaluating the successive integrals produced by the repeated integration by parts: √ √ x3 dx = 2x3 x + 1 − 3x2 2 x + 1 dx, 1/2 (x + 1) √ 2 2 x2 x + 1 dx = x2 (x + 1)3/2 − 2x (x + 1)3/2 dx, 3 3 2 2 x(x + 1)3/2 dx = x(x + 1)5/2 − (x + 1)5/2 dx, 5 5 2 (x + 1)5/2 dx = (x + 1)7/2 . 7 And so, remembering to carry forward the multiplicative factors generated at each stage, we have √ 16 32 x + 1 2x3 − 4x2 (x + 1) + x(x + 1)2 − (x + 1)3 + c 5 35 √ 2 x+1 5x3 − 6x2 + 8x − 16 + c. = 35 J= (b) Set x + 1 = u2 , giving dx = 2u du, to obtain J= =2 (u2 − 1)3 2u du u (u6 − 3u4 + 3u2 − 1) du.

They clearly do. (d) This is as in (c), but there are no real roots. However, it can be more generally stated that if there are two values of x that give 2x2 + 3x + k equal values then they lie one on each side of x = − 34 . 25 PRELIMINARY CALCULUS (e) With f(x) = 2x3 − 21x2 + 60x + k, f (x) = 6x2 − 42x + 60 = 6(x − 5)(x − 2) and f (x) = 0 has roots 2 and 5. Therefore, if f(x) = 0 has three real roots αi with α1 < α2 < α3 , then α1 < 2 < α2 < 5 < α3 . (f) When k = −45, f(3) = 54 − 189 + 180 − 45 = 0 and so x = 3 is a root of f(x) = 0 and (x − 3) is a factor of f(x).

35) and the subsequent text. 1 2 dt 2t 1 + t2 1+ 1 + t2 2 dt = (1 + t)2 2 +c =− 1+t 2 + c. =− θ 1 + tan 2 1 dθ = 1 + sin θ (d) To remove the square root, set u2 = 1 − x; then 2u du = −dx and 1 × −2u du (1 − u2 )u −2 = du 1 − u2 −1 −1 = + du 1−u 1+u = ln(1 − u) − ln(1 + u) + c √ 1− 1−x √ + c. 32 Express x2 (ax + b)−1 as the sum of powers of x and another integrable term, and hence evaluate b/a x2 dx. ax + b 0 We need to write the numerator in such a way that every term in it that involves x contains a factor ax + b.

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Instructor's solutions for Mathematical methods for physics and engineering, 3ed by Riley, Hobson.


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