The Natural Logarithm solved questions

CHAPTER 23
The Natural Logarithm

23.1 State the definition of In x, and show that D^(ln jc) =
_ !_
In* = dt for x > 0 Hence (Problem 20.42), D,(ln*)=D A x'


23.2 Show that dx = In 1*1 + C for x * 0

Case 1. x> 0. Then D,(ln|*| + C) = D,(ln*) = Case 2. x<0. Then D,(ln \x\ + C) =
D,[ln (-*)] = • D,(-x) = - -(-!) =

In Problems 23.3-23.9, find the derivative of the given function.

23.3 In (4* - 1).

By the chain rule, D,[ln(4*-l)] = •D,(4*-l) =

23.4 (In x)3.

By the chain rule, DA.[(ln x)3] = 3(ln x)- • Dx(\n x) = 3(ln x)2 •

23.5 VhT7.

By the chain rule, D^VHTx) = D,[(ln x)"2] = ^(Inx)'1'2 • Df(\nx)= ^Inx)'1'2 • - =

23.6 In (In*).

By the chain rule, DJln (In x)] = •D,(lnx) =

23.7 x2 In x.
By the product rule, Dx(x2 In x) = x2 • Dx(\n x) + In x • Dv(x2) = x2 • + In x • (2.v) = x + 2x In .v = .v(l +
2 In*).

23.8 In

By the chain rule,

In

23.9 ln|5*-2|.

By the chain rule, and Problem 23.2, D,(ln \5x - 2|) = •D,(5*-2) =

In Problems 23.10-23.19, find the indicated antiderivative.


23.10 dx.

dx = dx=$ \n\x\ + C.

185

, 186 CHAPTER 23


23.11 dx.

Let u = 7x-2, du=l dx. Then dx = du= % ln| M l + C = \\n\lx-2\ + C.


23.12 dx.

Let u = x4 — 1, du=4x*dx. Then dx = du = | In \u\ + C = | In \x4 - 1| + C.

23.13 J cot x dx.

Let u = sin x, du = cos x dx. Then J cot x dx = dx = dw = ln|w| + C = ln|sinA-| + C


23.14

Let w = In x, du = dx. Then dx = rf« = ln|w| + C = ln(|lnjc|)+C. [Compare Prob-
lem 23.6.]

23.15

Let w = 1 — sin 2x, du = —2 cos 2x dx Then

dx = du = -{ \n\u\ + C= -| I n | l - s i n 2 x | + C

23.16

rfx = 3 J A-2 rfx + 2 dx - 3 J x^ 3 dx = jc3 + 2 In |*| + l^r" 2 + C.


23.17

Let u - tan x, du = sec2 x dx. Then

dx = du = In |M| + C = In |tan x\ + C


23.18 dx.

Let u = \-Vx, du = -(l/2Vx)dx. Then

dx = -2 du = -2 In |«| + C = -2 In |1 - Vx\ + C


23.19 dx.

dx = (In*) dx={(lnx)2 + C.


23.20 Show that dx = \n\g(x)\ + C.

Let M = g(x), du = g'(x) dx. Then

dx = du = ln\u\ + C = \n\g(x)\ + C