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# It will be found on carrying out the substitution

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It will be found, on carrying out the substitution, that the integral (5) as- sumes the form H Z t dt ( αt 2 + β ) p γt 2 + δ + K Z dt ( αt 2 + β ) p γt 2 + δ . (6) The second of these integrals is rationalised by the substitution t p γt 2 + δ = u,

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[VI : 139] DERIVATIVES AND INTEGRALS 296 which gives Z dt ( αt 2 + β ) p γt 2 + δ = Z du β + ( αδ - βγ ) u 2 . Finally, if we put t = 1 /u in the first of the integrals (6), it is transformed into an integral of the second type, and may therefore be calculated in the manner just explained, viz. by putting u/ p γ + δu 2 = u , i.e. 1 / p γt 2 + δ = v . * Examples L. 1. Evaluate Z dx x x 2 + 2 x + 3 , Z dx ( x - 1) x 2 + 1 , Z dx ( x + 1) 1 + 2 x - x 2 . 2. Prove that Z dx ( x - p ) p ( x - p )( x - q ) = 2 q - p r x - q x - p . 3. If ag 2 + ch 2 = - ν < 0 then Z dx ( hx + g ) ax 2 + c = - 1 ν arc tan " p ν ( ax 2 + c ) ch - agx # . 4. Show that Z dx ( x - x 0 ) y , where y 2 = ax 2 + 2 bx + c , may be expressed in one or other of the forms - 1 y 0 log axx 0 + b ( x + x 0 ) + c + yy 0 x - x 0 , 1 z 0 arc tan axx 0 + b ( x + x 0 ) + c yz 0 , according as ax 2 0 + 2 bx 0 + c is positive and equal to y 2 0 or negative and equal to - z 2 0 . * The method of integration explained here fails if a/A = b/B ; but then the integral may be reduced by the substitution ax + b = t . For further information concern- ing the integration of algebraical functions see Stolz, Grundz¨uge der Differential-und- integralrechnung , vol. i, pp. 331 et seq. ; Bromwich, Elementary Integrals (Bowes and Bowes, 1911). An alternative method of reduction has been given by Sir G. Greenhill: see his A Chapter in the Integral Calculus , pp. 12 et seq. , and the author’s tract quoted on p. 286 .
[VI : 140] DERIVATIVES AND INTEGRALS 297 5. Show by means of the substitution y = ax 2 + 2 bx + c/ ( x - p ) that Z dx ( x - p ) ax 2 + 2 bx + c = Z dy p λy 2 - μ , where λ = ap 2 + 2 bp + c , μ = ac - b 2 . [This method of reduction is elegant but less straightforward than that explained in § 139 .] 6. Show that the integral Z dx x 3 x 2 + 2 x + 1 is rationalised by the substitution x = (1 + y 2 ) / (3 - y 2 ). ( Math. Trip. 1911.) 7. Calculate Z ( x + 1) dx ( x 2 + 4) x 2 + 9 . 8. Calculate Z dx (5 x 2 + 12 x + 8) 5 x 2 + 2 x - 7 . [Apply the method of § 139 . The equation satisfied by μ and ν is ξ 2 + 3 ξ + 2 = 0, so that μ = - 2, ν = - 1, and the appropriate substitution is x = - (2 t + 1) / ( t + 1). This reduces the integral to - Z dt (4 t 2 + 1) 9 t 2 - 4 - Z t dt (4 t 2 + 1) 9 t 2 - 4 . The first of these integrals may be rationalised by putting t/ 9 t 2 - 4 = u and the second by putting 1 / 9 t 2 - 4 = v .] 9. Calculate Z ( x + 1) dx (2 x 2 - 2 x + 1) 3 x 2 - 2 x + 1 , Z ( x - 1) dx (2 x 2 - 6 x + 5) 7 x 2 - 22 x + 19 . ( Math. Trip. 1911.) 10. Show that the integral Z R ( x, y ) dx , where y 2 = ax 2 + 2 bx + c , is ra- tionalised by the substitution t = ( x - p ) / ( y + q ), where ( p, q ) is any point on the conic y 2 = ax 2 + 2 bx + c . [The integral is of course also rationalised by the substitution t = ( x - p ) / ( y - q ): cf. § 134 .]

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[VI : 141] DERIVATIVES AND INTEGRALS 298 140. Transcendental Functions. Owing to the immense variety of the different classes of transcendental functions, the theory of their in- tegration is a good deal less systematic than that of the integration of rational or algebraical functions. We shall consider in order a few classes of transcendental functions whose integrals can always be found.
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