An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 120-121.

Reduction to rational forms.

203

Thus we have:
Vq- C.
-=tan —_=t + C0
I/~-c (x - a)- -c
These substitutions can also be employed from the very beginning
to convert
JF(x, 1/W)dx
into the integral of a rational function.
Note. The case c — = 0 reduces to the simplest binomial integral,
the integral then becomes:
fa4 d1x ->- /a +- 2bx + C.
121. In order to determine the integral:
li" dx
let us begin by investigatingj /-x; we find a recurring formula for
it from:
z yhtS0 - 1 d (7 b+c )dx
\x- n1 xn x~-I n cVI
- - 1 (a - 2bx - -cx2) (b -c- cx) d
- ~- -_ V_ C_ _+-_1- dV
(n a- 1)adx (2n - 3)bdX (n -- 2)cdx
-- x /-1 - V 1/ x~n-2f;
thus if a 0:
Iaf / I - _1 /R 22n-3 b ( dx I-S _ cj' dx
TJ xG y (n - 1)x - a x - -, a x - -1 J - -- J xS 2 1/
When we put for n the values 2, 3, 4,... in succession, the integral
reduces ultimately to the form j dx
The formula is generalised by substituting x = - e; then,
R == A + 2Bs + C'- -2 (a - 2 b + c2) + 2(b - c)Z -+ csz2
thus:
c C, b = B + Co, a =- A + 2BQ + CQ2.
Now write on both sides a, b, c, for A, B, C, and for z put x,
also for brevity write:
C + 2bQ + ce-/(Q ) + = v-/n(q), b /"(),
then, provided /'(9) does not vanish, we have: