﻿ Natural Logarithmic Integration Worksheet | Problems & Solutions

# Natural Logarithmic Integration Worksheet

Natural Logarithmic Integration Worksheet
• Page 1
1.
If = ln ( ||) + C,
then evaluate .
 a. 6 $x$ + C b. - 24 $x$ + C c. 4 $x$ + C d. 24 $x$ + C

#### Solution:

x² +6x- 33x³ - 4x² - 9x + 36dx = x² +6x-33(x + 3)(x - 3)(x - 4)dx
[Factor the denominator of the original rational function.]

= (-1x + 3 +1x - 3 +1x - 4)dx
[Resolve the rational function into partial fractions.]

= - ln | x + 3| + ln | x - 3| + ln | x - 4| + C
[Use 1x+adx = ln |x + a| + C.]

= ln (| x + 3|)-1 + ln | x - 3| + ln | x - 4 | + C
[Use k ln a = ln ak.]

= ln( | (x - 4)(x - 3)x +3|) + C
[Use ln a + ln b + ln c = ln (abc).]

k = 4
[Compare with ln ( |(x - k)(x - 3)x + 3]|) + C to get the value of k.]

6 k dx = 24 dx = 24 x + C
[Evaluate 6 k dx.]

2.
Evaluate:
 a. + C b. + C c. + C d. + C

#### Solution:

(5 + ln 24x)²xdx = u² du
[Let u = 5 + ln 24x. Then du = (1x) dx.]

= (u³3) + C
[Use xkdx =xk+1k+1 + C, for k ≠ -1.]

= (5 + ln 24x)³3 + C
[Substitute u = 5 + ln 24x.]

3.
Evaluate:
 a. 43 ln |7 + 43 $x$| + C b. $\frac{3}{43}$ ln |7 + 43 $x$| + C c. $\frac{3}{43}$ ln |7 - 43 $x$| + C d. 7 ln |7 + 43 $x$| + C

#### Solution:

37 + 43xdx = 3 17 + 43xdx
[Use k f(x) dx = kf(x) dx.]

= 3 / 431udu
[Let u = 7 + 43x. Then du = 43 dx.]

= 3 / 43 ln |u| + C
[Use 1xdx = ln |x| + C.]

= 3 / 43 ln |7 + 43x| + C
[Substitute u = 7 + 43x.]

4.
Evaluate:
 a. + C b. + C c. - + C d. + C

#### Solution:

ln (33x)xdx = udu
[Let ln (33x) = u. Then (1x) dx = du.]

= (u²2) + C
[Use xkdx =xk+1k+1 + C, for k ≠ -1.]

= (ln (33x))²2 + C
[Substitute ln (33x) = u.]

5.
If = A ln | 4$x$ + 9 | + C , then find A.
 a. $\frac{1}{4}$ b. $\frac{1}{16}$ c. $\frac{1}{9}$ d. $\frac{1}{81}$

#### Solution:

dx16x² + 72x + 81 = dx 4x + 9
[Factor the expression in the radical sign of the denominator and simplify.]

= 14 ln |4x + 9| + C
[Use dx ax + b  = 1a ln |ax + b| + C.]

The value of A is 14.
[Compare 14 ln |4x + 9| + C with A ln | 4x + 9 | + C.]

6.
Evaluate:
 a. ln |$e$12$t$ + 2| + C b. ($\frac{1}{12}$) ln |$e$$t$ + 2| + C c. ($\frac{1}{12}$) ln |$e$12$t$ - 2| + C d. ($\frac{1}{12}$) ln |$e$12$t$ + 2| + C

#### Solution:

e12te12t + 2dt = (1 / 12) 1udu
[Let u = e12t + 2. Then du = 12e12t dt.]

= (1 / 12) ln |u| + C
[Use 1xdx = ln |x| + C.]

= (1 / 12) ln |e12t + 2| + C
[Substitute u = e12t + 2.]

7.
Evaluate:
 a. $x$ ln (17ln $x$) + x (ln $x$)-1 + C b. - $x$ ln (18ln $x$) - (ln $x$)-1 + C c. $x$ ln (16ln $x$) - $x$ (ln $x$)-1 + C d. $x$ ln 16$x$ + C

#### Solution:

(ln(16ln x)+(ln x)-2)dx = x ln (16ln x) - xxln xdx + (ln x)-2dx + C
[Use Integration by parts.]

= x ln (16ln x) - (ln x)-1dx + (ln x)-2dx + C
[Rewrite the expression.]

= x ln (16ln x) - [x (ln x)-1 + (ln x)-2dx] + (ln x)-2dx + C
[Use Integration by parts again.]

= x ln (16ln x) - x (ln x)-1 + C
[Simplify.]

8.
If | + C, then choose $f$($x$) from the following.
 a. - ${\mathrm{Tan}}^{-1}\frac{x}{4}$ b. 3 ${\mathrm{Tan}}^{-1}\frac{x}{4}$ c. 2${\mathrm{Tan}}^{-1}\frac{x}{4}$ d. ${\mathrm{Tan}}^{-1}\frac{x}{4}$

#### Solution:

x16+x²dx=121tdt
[Let 16 + x2 = t Þ 2xdx = dt.]

= 1 / 2 t-12dt=t12
[Use xkdx =xk+1k+1 + C, for k ≠ -1.]

= t =16+x2
[Substitute t = 16 + x2.]

xtan-1x416 + x²dx = (Tan-1x4)16+x²-416+x²16+x²dx
[Use Integration by parts.]

= 16 + x²(Tan-1x4) - 416 + x²dx
[Simplify.]

= 16 +x²(Tan-1x4) - 4 ln|x + 16+x²| + C
[Use dxa²+x² = ln |x+a²+x²|.]

Thus f(x) = Tan-1x4
[Compare 16 +x²(Tan-1x4) - 4 ln|x + 16+x²| + C with 16 + x²f(x) - 4 ln |x + 16 + x²| + C.]

9.
Evaluate:
 a. + C b. + C c. + C d. + C

#### Solution:

e24x(2 + sin 48x)1 + cos 48x = e24x[2 + 2 sin (24x) cos (24x)]2cos2(24x)dx
[Use half angle formulas.]

= e24x[sec² 24x + tan 24x]dx
[Divide both the numarator, the denominator by 2 cos224x.]

= e24x sec² 24x dx + e24x tan 24xdx
[Use (u+v) dx =u dx +v dx.]

= e24x sec² 24x dx + e24x24 tan 24x - e24x sec² 24xdx + C
[Use integration by parts for e24x sec² 24x dx.]

= e24x  tan 24x24 + C
[Simplify.]

10.
If I = = A cos 5$x$ + B ln |$f$($x$)| + C, then which of the following is correct.
 a. A = - 4, B = - $\frac{6}{\sqrt{2}}$ b. A = 4, B = - $\frac{6}{\sqrt{2}}$ c. A = - 4, B = $\frac{6}{\sqrt{2}}$ d. A = 4, B = $\frac{6}{\sqrt{2}}$

#### Solution:

40(sin 5x + sin³ 5xcos 10x)dx = 40 sin 5x(1 + sin² 5x)cos 10xdx
[Factor the expression in the numerator and use k f(x) dx = kf(x) dx.]

= 40 sin 5x(1 + (1-cos² 5x))cos 10xdx
[Use sin2 a = 1 - cos2 a.]

= 40 sin 5x(2 - cos² 5x)cos 10xdx
[Simplify.]

= 8 t² - 22t² - 1dt
[Let cos 5x = t, then (- 5 sin 5x) dx = dt and then use the half angle formuls.]

= 4 dt - 12 dt2t² - 1
[Resolve the rational function into partial fractions.]

= 4 t - (1222) ln |(2t - 1)(2t + 1)| + C
[Use dt(at)2 - 1 =12a ln |at - 1at + 1| + C.]

= 4 cos 5x - 62 ln|(2 cos 5x - 1)(2 cos 5x + 1)| + C
[Substitute t = cos 5x.]

A = 4 , B = - 62
[Compare 4 cos 5x - 122 ln|(2 cos 5x - 1)(2 cos 5x + 1)| + C with A cos 5x + B ln |f(x)| + C and write the values of A, B.]