146 PRACTICAL MATHEMATICS 



Case I. When the denominator is the product of unlike linear 

 factors. 



/ \ r i 5# + 4 



(a) To integrate ; -r-. -r 



(x- 2)(x + 5) 



XT 5x + 4 A B 



Now ; - ; r = + 



(x-2)(x+ 5) x-2 x+ 5 

 and A(x + 5) + B (x - 2) = 5x + 4 



When a?=27A =14 A = 2 



When a? = 5 -7B =-21 B = 3 



Then f, (te + 4)A L - af-*- + 8f-t= 



J (# J ) (a; + 5) jx & jx + 5 



(&) To integrate 



= 2 log c (a; - 2) + 3 log e (x + 5) 

 x+ 8 



(5x- 3) (4- to) 



x+ 8 A B 



Now - = + 



(Be - 3)(4 - 3x) 5x-3 4,-3x 

 and A(4 - 3a?) + B(5 - 3) = x + 



,, 3 11 . 43 43 



When a.= - y A -- A = - 



4 11 28 28 



When a, = - y B = - B = n 



, r (a; + 8) dx 43j* <fe 28f (fe 



J(5a? - 3)(4 - 3x) ~ lljBx - 3 + TTJ4 - ; 



43P 5 (to 28f-3 



" 55j5ir-3~33j 4- 



3aj 



-3 dx 

 ~3x 



It should be noticed in this example that when integrating 

 the partial fractions, the numerators are not the differential 

 coefficients of the denominators, but by multiplying and dividing 

 the first fraction by 5 and the second fraction by 3 this relation 

 is at once made to hold. 



Case II. When the denominator of the fraction consists of one 

 linear factor raised to a power. 



T, . , 5^ 



To integrate 



Putting y = 2x - 3, then x = -(y + 3) 



m 



575 



and 5x z - 8x + 2 = -gf + -^ + - 



