9.5 Integration by parts 521We could feel that integration by parts would not enableus to simplify
f log x dx, but we can set
u=logx, dv=ldx
du=1dx, v=x
xto obtain
f logxdx=xlogx- f1dx=xlogx-x+c.
This result is easily checked by differentiation.
Problems 9.59
1 Derive the following formulas by one or more integrations by parts(a) f xea: dx =(x-^1
(a az) a°x + c
x2 x2
(b) f x log x dx = 2 log x- 4+ c
xn+l 1
(c) f x" log x dx =
;T_+_1(logx - n +1) -{-c
(d) f x2eaa dx = (-X^2 - 2 + a3) eaz + c2x^2
' a a2(e) f xsec2xdx=xtanx+logcosx+c
(f) f (log x)n dx = x(log x)n - n f (log x)n-1 dx(g) f(x+1), dx = z1+1 log (x -} i) + c
(h) f sin-' xdx =xsin'x-1 1 -x2+c
(i) f tan' x dx = x tan' x - log (1 + x2) + c
2 Setting u = iP, dv = (1 - t)4 dt, derive the formulafol i (1 - t)4 dt =q+^1
fo1 t-1(1 - t)4+1 dt.Observe the fact that the result agrees with the beta integral formula
i
fo tP(1 - t)4 dtp !q!
= (p +q + 1)1and the formula r! = r[(r - 1)!]. Remark: In Problems 18 and 19 of Problems
13.49, we shall introduce the Euler gamma integral formulaz! = fo0 t=e t dt (z > -1)
and show how it is used to derive the Euler beta integral formula.