The measure of minor arc JL is 60°. Circle M is shown. Line segments M J and M L are radii. Tangents J K and L K intersect at point K outside of the circle. Arc J L is 60 degrees. What is the measure of angle JKL? 110° 120° 130° 140°

Power and chain rule (where the power rule kicks in because [tex]\sqrt x=x^{1/2}[/tex]):

[tex]\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'[/tex]

Simplify the leading term as

[tex]\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}[/tex]

Quotient rule:

[tex]\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}[/tex]

Chain rule:

[tex](\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)[/tex]

[tex](1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)[/tex]

Put everything together and simplify:

[tex]\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}[/tex]

[tex]=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}[/tex]

[tex]=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}[/tex]

[tex]=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}[/tex]

[tex]=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}[/tex]

[tex]=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}[/tex]

[tex]=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}[/tex]

Answer:

[tex] E(X) =\int_{4}^7 \frac{1}{3} x[/tex]

[tex] E(X) = \frac{1}{6} (7^2 -4^2) = 5.5[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) =\int_{4}^7 \frac{1}{3} x^2[/tex]

[tex] E(X^2) = \frac{1}{9} (7^3 -4^3) = 31[/tex]

And the variance for this case would be:

[tex] Var(X)= E(X^2) -[E(X)]^2 = 31 -(5.5)^2 = 0.75[/tex]

And the standard deviation is:

[tex] Sd(X)= \sqrt{0.75}= 0.866[/tex]

Step-by-step explanation:

For this case we have the following probability density function

[tex] f(x)= \frac{1}{3}, 4 \leq x \leq 7[/tex]

And for this case we can find the expected value with this formula:

[tex] E(X) =\int_{4}^7 \frac{1}{3} x[/tex]

[tex] E(X) = \frac{1}{6} (7^2 -4^2) = 5.5[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) =\int_{4}^7 \frac{1}{3} x^2[/tex]

[tex] E(X^2) = \frac{1}{9} (7^3 -4^3) = 31[/tex]

And the variance for this case would be:

[tex] Var(X)= E(X^2) -[E(X)]^2 = 31 -(5.5)^2 = 0.75[/tex]

And the standard deviation is:

[tex] Sd(X)= \sqrt{0.75}= 0.866[/tex]

Answer:

Step-by-step explanation:

3(2)^2 + 4(1)^2

3(4) + 4

12+4= 16

Answer:

[tex]\huge\boxed{16}[/tex]

Step-by-step explanation:

[tex]3x^2+4y^2\ \text{for}\ x=2;\ y=1.\\\\\text{Substitute:}\\\\3(2)^2+4(1)^2=3(4)+4(1)=12+4=16\\\\\text{Used PEMDAS}[/tex]

Answer:

(x – 4)^2 + (y + 7)^2 = 16

Step-by-step explanation:

The formula of a circle is:

(x – h)^2 + (y – k)^2 = r^2

(h, k) represents the coordinates of the center of the circle

r represents the radius of the circle

If you plug in the given information, you get:

(x – 4)^2 + (y – (-7))^2 = 4^2

which simplifies into:

(x – 4)^2 + (y + 7)^2 = 16

The answer is 86.4 km

Explanation:

The graph shows the gift shop is to the east of the science lab, and, between the gift shop and the science lab it is the art supply. Besides this, the description of the graph provides the distance between the art supply and the science lab, which is 40.0, as well as, the distance between the art supply and the gift shop, which is 46.4 kilometers.

In this context, it is possible to calculate the distance from the science lab to the gift shop by adding the partial distances, considering the art supply as a middle point in the map. This means the distance from the lab to the gift shop = 40.0 km (distance from the lab to the art supply) + 46.4 km (distance from the art supply to the gift shop) = 86.4 km.

Answer:

1.35

Step-by-step explanation:

next cube above 540 is 729

to get to 729: 729 / 540 = 1.35

n = 1.35

Answer:

y=0.5x+40

Step-by-step explanation:

Copy the  equation.

x-2y=8

Subtract x from both sides.

-2y=-x-8

Divide both sides by -2.

y=0.5x+4

Now we know the slope is 0.5.

Any line with a slope of 0.5 will be perpendiculr to the original line.

One that you can use is y=0.5x+40.

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = Number of adults who use their smartphones

The above situation can be represented through the binomial distribution;

[tex]P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......[/tex]

where, n = number of trials (samples) taken = 15 adult smartphones

           r = number of success = at least 4

           p = probability of success which in our question is the % of adults

                 who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X [tex]\geq[/tex] 4)

P(X [tex]\geq[/tex] 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= [tex]1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}[/tex]

= [tex]1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})[/tex]

= 0.1773

Answer:

3

Step-by-step explanation:

3

Answer: 0.0548

Step-by-step explanation:

Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.

Let [tex]\overline{X}[/tex] represents the sample mean GPA for each student.

Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:

[tex]P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z<1.6)\\\\=1-0.9452=0.0548[/tex]

hence, the required probability is 0.0548.

Answer:

1584$

Step-by-step explanation:

Original price is 1800$ (100%)

Discount percent: 12%

=> The price after discount is 100 - 12 = 88% of original price

=> The price after discount is 1800 x 88% = 1800 x 88/100 = 1584$

Answer:

13200

Step-by-step explanation:

12% - 1800

100% - x

X = (1800x100)/12 = 15000 - original price

15000-1800 = original price - discount = 13200 price after discount

Answer:

Option D

Step-by-step explanation:

A type I error occurs when you reject the null hypothesis when it is actually true.

The null hypothesis in this case is minimum breaking strength is less than or equal to 0.5.

A type one error would be allowing the production process to continue when the true breaking strength is below specifications.

Answer:

linear

Step-by-step explanation:

Answer:

see explanation

Step-by-step explanation:

The median is the middle value of the data set in ascending order. If there is no exact middle then the median is the average of the values either side of the middle.

Given

3   8   14   19   22   29   33   37   43   49

                            ↑ middle is between 22 and 29

median = [tex]\frac{22+29}{2}[/tex] = [tex]\frac{51}{2}[/tex] = 25.5

The upper quartile [tex]Q_{3}[/tex] is the middle value of the data to the right of the median.

29   33   37   43   49

               ↑

[tex]Q_{3}[/tex] = 37

The lower quartile [tex]Q_{1}[/tex] is the middle value of the data to the left of the median.

3   8   14   19   22

           ↑

[tex]Q_{1}[/tex] = 14

The min is the smallest value in the data set, that is 3

The max is the largest value in the data set, that is 49

The 5 number summary is

3,  14,  25.5,  37,  49

Answer:

Step-by-step explanation:

Hello,

First of all, we need to find the equation of the line.

It will be something like y = ax + b (we need to find a and b).

Then the graph is the part of the plan which is above this line, so the inequality will be

[tex]\large \boxed{\sf \ \ y\geq ax+b \ \ }[/tex]

There is only one line passing by two different points, right?

We need to find two points of this line, the x-axis gives the x of the point and the y-axis give the y of the point.

We can see on the graph that (-1,-1) and (6,13) are two points of this line.

I attached a graph with the two points A (-1,-1) and B (6,13).

We need to solve y = ax + b to find a and b which means:

(-1) = a(-1 )+ b <=> -1 = -a + b <=> -a + b = -1

(13) = a(6) + b <=> 6a + b = 13

To eliminate b, we can subtract the first equation from the second one.

6a + b -(-a) - b = 13 - (-1) = 13 + 1 = 14

<=> 6a + a = 7a = 14 we can divide by 7 both parts of the equation.

[tex]\large \boxed{\sf \ \ a=\dfrac{14}{7}=2 \ \ }[/tex]

And then, b = -1 + a = -1 + 2 = 1

[tex]\large \boxed{\sf \ \ b=1 \ \ }[/tex]

The equation of the line is then [tex]y=2x+1[/tex]

So, the inequality is:

[tex]\Large \boxed{\sf \ \ y\geq 2x+1 \ \ }[/tex]

And this is the answer of the question 1.

2. First, we find the equation of the line by finding two points on this line. Then, we deduce the inequality from the equality as the graph is all the points above the line (as explained above).

3. A real-life example is the following.

My parents gave $1 to my sister and they will give her $2 every day.

So today at x = 0 she has $1 and her pocket money will be represented by the line. And I told my parents that there were no way I can get less than my little sister. To help them understand what can be an acceptable deal I provided this graph, and I told them to only look for t positive of course.

The label for the x-axis is the time.

The label for the y-axis is the pocket money in $

After one day x = 1 my sister has $3 (point on the line), a possible solution for my pocket money is $6, $14 (two points from the graph) is also a solution of course :-)

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer: [tex]S=0.087p[/tex] .

Step-by-step explanation:

Equation for direct proportion:

y=kx

, where x= independent variable ,

y=dependent variable.

k= proportionality constant

Here, State sales tax S  is directly proportional to retail price p.

Also, dependent variable= S,  independent variable =p

Required equation: S= kp

Put S= 12.32 and x= 142

[tex]S=12.32=k(142)\\\\\Rightarrow\ k=\dfrac{12.32}{142}\approx0.087[/tex]

Hence, the required equation is [tex]S=0.087p[/tex] .

Answer:

y-intercept is (0,-10) and x-intercept is (-6,0).  Connect them by a straight line to graph the given equation.

Step-by-step explanation:

The given equation of line is

[tex]3y=-5x-30[/tex]

For x=0,

[tex]3y=-5(0)-30[/tex]

[tex]3y=-30[/tex]

[tex]y=-10[/tex]

So, y-intercept is at point (0,-10).

For y=0,

[tex]3(0)=-5x-30[/tex]

[tex]0=-5x-30[/tex]

[tex]5x=-30[/tex]

[tex]x=-6[/tex]

So, x-intercept is at point (-6,0).

Now, plot the point (0,-10) and (-6,0) on a coordinate plane and connect them by a straight line to graph the given line as shown below.

Answer:

[tex]\large \boxed{\sf \text{The rate of the boat is } 53 \ km/h \text{, the rate of the current is }8\ km/h \ \ }[/tex]

Step-by-step explanation:

Hello, let's note v the rate of the boat and r the rate of the current. We can write the following

[tex]\dfrac{135}{v-r}=3=\dfrac{183}{v+r}[/tex]

It means that

[tex]135(v+r)=183(v-r)\\\\135 v + 135r=183v-183r\\\\\text{ *** We regroup the terms in v on the right and the ones in r to the left***}\\\\(135+183)r=(183-135)v\\\\318r=48v\\\\\text{ *** We divide by 48 both sides ***}\\\\\boxed{v = \dfrac{318}{48} \cdot r= \dfrac{159}{24} \cdot r}[/tex]

But we can as well use the second equation:

[tex]3(v+r)=183\\\\v+r=\dfrac{183}{3}=61\\\\\dfrac{159}{24}r+r=61\\\\\dfrac{159+24}{24}r=61\\\\\boxed{r = \dfrac{61*24}{183}=8}[/tex]

and then

[tex]\boxed{v=\dfrac{159*8}{24}=53}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Question:

Vector A has x and y components of −8.80 cm  and 18.0 cm , respectively; vector B has x and  y components of 12.2 cm and −6.80 cm , respectively.  If A − B +3 C = 0, what are the components of C?

Answer:

x = ___ cm

y = ___ cm

Answer:

x = 7.0cm

y = -8.27cm

Step-by-step explanation:

For a vector F, with x and y components of a and b respectively, its unit vector representation is as follows;

F = ai + bj              [Where i and j are unit vectors in the x and y directions respectively]

Using this analogy, let's represent vectors A and B from the question in their unit vector notation.

A has an x-component of -8.80cm and y-component of 18.0cm

B has an x-component of 12.2cm and y-component of -6.80cm,

In unit vector notation, these become;

A = -8.80i + 18.0j

B = 12.2 i + (-6.80)j = 12.2i - 6.80j

Also, there is a third vector C. Let the x and y components of C be a and b respectively. Therefore,

C = ai + bj

Now,

A - B + 3C = 0                [substitute the vectors]

=> [-8.80i + 18.0j] - [12.2 i -6.80j] + [3(ai + bj)] =  0        [open brackets]

=> -8.80i + 18.0j - 12.2 i + 6.80j + 3(ai + bj) =  0

=> -8.80i + 18.0j - 12.2 i + 6.80j + 3ai + 3bj =  0

=> -8.80i + 18.0j - 12.2 i + 6.80j + 3ai + 3bj =  0  [collect like terms and solve]

=> -8.80i  - 12.2 i  + 3ai + 6.80j + 18.0j + 3bj =  0

=> -21.0 i  + 3ai + 24.8j + 3bj =  0       [re-arrange]

=> 3ai + 3bj = 21.0i - 24.8j

Comparing both sides shows that;

3a = 21.0  -------------(i)

3b = -24.8    -----------(ii)

From equation (i)

3a = 21.0

a = 21.0 / 3 = 7.0

From equation (ii)

3b = -24.8

b = -24.8 / 3

b = -8.27

Therefore, the x-component and y-component of vector B which are a and b, are 7.0cm and -8.27cm respectively.

Answer:  203,280

Step-by-step explanation:

Given: A catering service offers 11 appetizers, 12 main courses, and 8 desserts.

Number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

A customer is to select 9 appetizers, 2 main courses, and 3 desserts for a banquet.

Total number of ways to do this: [tex]^{11}C_9\times ^{12}C_2\times^{8}C_3[/tex]

[tex]=\dfrac{11!}{9!2!}\times\dfrac{12!}{2!10!}\times\dfrac{8!}{3!5!}\\\\=\dfrac{11\times10}{2}\times\dfrac{12\times11}{2}\times\dfrac{8\times7\times6}{3\times2}\\\\= 203280[/tex]

hence , this can be done in 203,280 ways.

Answer:

$10 to rent shoes for 9 people

Step-by-step explanation:

Total amount of the party = $140

A bowling lane = $50

$140 - $50 = $90

$90 divided by 9 = 10

$10 to rent shoes for 9 people

Answer:

-1.031 m/s or  [tex]\frac{-\sqrt{17} }{4}[/tex]

Step-by-step explanation:

We take the length of the rope from the dock to the bow of the boat as y.

We take x be the horizontal  distance from the dock to the boat.

We know that the rate of change of the rope length is [tex]\frac{dy}{dt}[/tex] = -1 m/s

We need to find the rate of change of the horizontal  distance from the dock to the boat =  [tex]\frac{dx}{dt}[/tex] = ?

for x = 4

Applying Pythagorean Theorem we have

[tex]1^{2} +x^{2} =y^{2}[/tex]    .... equ 1

solving, where x = 4, we have

[tex]1^{2} +4^{2} =y^{2}[/tex]

[tex]y^{2} = 17[/tex]

[tex]y = \sqrt{17}[/tex]

Differentiating equ 1 implicitly with respect to t, we have

[tex]2x\frac{dx}{dt} = 2y\frac{dy}{dt}[/tex]

substituting values of

x = 4

y = [tex]\sqrt{17}[/tex]

[tex]\frac{dy}{dt}[/tex] = -1

into the equation, we get

[tex]2(4)\frac{dx}{dt} = 2(\sqrt{17} )(-1)[/tex]

[tex]\frac{dx}{dt} = \frac{-\sqrt{17} }{4}[/tex] = -1.031 m/s

Answer:

3/11

Step-by-step explanation:

There are eleven equal parts.

So the denominator is 11.

He copies 8 parts on Sunday.

11-8=3.

He copied 3 parts on Saturday.

Hope this helps ;) ❤❤❤

Answer:

"B. about half the days in march had average temperatures either below 60 or above 73 degrees"

Step-by-step explanation:

To answer this question, note that a box plot is usually divided into quartiles, each representing approximately 25% each.

In the box plot above,

*about 25% (Q1) represents days with temperature of 60° and below. This is about ¼ of the days in March.

*About 25% (Q2) represents days with temperature between 61° and 68°. That's about ¼ of the days in March

*About 25% (Q3) represents days with temperature between 70° and 73°. That's about ¼ of the days in March

*About 25% (Q4) represents days with temperature between 74° and 82°. That's about ¼ of the days in March

*Coldest day in March has a temperature of 54°

*Hottest day in March is 82°

From the options given, the only statement that is true is "B. about half the days in march had average temperatures either below 60 or above 73 degrees"

¼ of the Days in March has temperatures below 60° (Q1), while ¼ of the days in March has temperatures above 73° (Q4). Therefore, ¼+¼ = ½ of the days in March having average temperatures either below 60 or above 73 degrees.

Answer:

b

Step-by-step explanation:

About half of the days in March had average temperatures either below 60 or above 73 degrees.

Answer:

[tex]\boxed{\sf x - 5 = 10}[/tex]

Step-by-step explanation:

Taking away 5 from x ⇒ subtracting 5 from x

[tex]\large {\sf x - 5[/tex]

Gives 10 ⇒ result is 10

[tex]\large {\sf x - 5 = 10[/tex]

Answer: 6 / √26

Step-by-step explanation:

Given that f(x, y, z) = xe^y + ye^z + ze^x

so first we compute the gradient vector at (0, 0, 0)

Δf ( x, y, z ) = [ e^y + ze^x,  xe^y + e^z,  ye^z + e^x ]

Δf ( 0, 0, 0 ) = [ e⁰ + 0(e)⁰, 0(e)⁰ + e⁰, 0(e)⁰ + e⁰ ] = [ 1+0 , 0+1, 0+1 ] = [ 1, 1, 1 ]

Now we were also given that  V = < 4, 3, -1 >

so ║v║ = √ ( 4² + 3² + (-1)² )

║v║ = √ ( 16 + 9 + 1 )

║v║ = √ 26

It must be noted that "v"  is not a unit vector but since ║v║ = √ 26, the unit vector in the direction of "V" is ⊆ = ( V / ║v║)

so

⊆ =  ( V / ║v║) = [ 4/√26, 3/√26, -1/√26 ]

therefore by equation   D⊆f ( x, y, z ) = Δf ( x, y, z ) × ⊆

D⊆f ( x, y, z ) = Δf ( 0, 0, 0 ) × ⊆ = [ 1, 1, 1 ] × [ 4/√26, 3/√26, -1/√26 ]

= ( 1×4 + 1×3 -1×1 ) / √26

= (4 + 3 - 1) / √26

= 6 / √26

Answer:

  28, 30, 32

Step-by-step explanation:

Their average will be 90/3 = 30. That is the middle integer.

The three integers are 28, 30, 32.

_____

Comment on the working

It often works well to use the average value when working consecutive integer problems. The average of an odd number of consecutive integers is the middle one. The average of an even number of consecutive integers is halfway between the middle two.

Answer

Step-by-step explanation:

Answer:

See the argument below

Step-by-step explanation:

I will give the argument in symbolic form, using rules of inference.

First, let's conclude c.

(1)⇒a  by simplification of conjunction

a⇒¬(¬a) by double negation

¬(¬a)∧(2)⇒¬(¬c) by Modus tollens

¬(¬c)⇒c by double negation

Now, the premise (5) is equivalent to ¬d∧¬h which is one of De Morgan's laws. From simplification, we conclude ¬h. We also concluded c before, then by adjunction, we conclude c∧¬h.

An alternative approach to De Morgan's law is the following:

By contradiction proof, assume h is true.

h⇒d∨h by addition

(5)∧(d∨h)⇒¬(d∨h)∧(d∨h), a contradiction. Hence we conclude ¬h.  

Answer:

A) 0.99413

B) 0.00022

Step-by-step explanation:

A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:

Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes

Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:

μ = n*μ_s ample = 42 × 5 = 210 minutes

While the standard deviation for the population would be:

σ = √nσ_sample = √(42 × 6) = 15.8745 minutes

To find the z-score, we will use the formula;

z = (x - μ)/σ

Thus;

z = (250 - 210)/15.8745

z = 2.52

From the z-distribution table attached, we have;

P(Z < 2.52) ≈ 0.99413

B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.

Thus, total time is now 250 + 10 = 260 minutes

Similar to the z-formula in A above, we have;

z = (260 - 210)/15.8745

z = 3.15

P(Z > 3.15) = 0.00022

Your integrand is missing some symbols. My best interpretation is the following integral:

[tex]I=\displaystyle\int\frac{x^4+18x^2+4}{x^5+30x^3+20x}\,\mathrm dx[/tex]

Decompose into partial fractions; we're looking for an expansion of the form

[tex]\dfrac{x^4+18x^2+4}{x^5+30x^3+20x}=\dfrac ax+\dfrac{bx^3+cx^2+dx+e}{x^4+30x^2+20}[/tex]

Now:

[tex]x^4+18x^2+4=a(x^4+30x^2+20)+(bx^3+cx^2+dx+e)x[/tex]

[tex]=(a+b)x^4+cx^3+(30a+d)x^2+ex+20a[/tex]

Matching up coefficients tells us that

[tex]\begin{cases}a+b=1\\c=0\\30a+d=18\\e=0\\20a=4\end{cases}\implies a=\dfrac15,b=\dfrac45,d=12[/tex]

so that

[tex]I=\displaystyle\frac15\int\frac{\mathrm dx}x+\frac45\int\frac{x^3+15x}{x^4+30x^2+20}\,\mathrm dx[/tex]

The integral is trivial:

[tex]\displaystyle\frac15\int\frac{\mathrm dx}x=\frac15\ln|x|+C[/tex]

For the second integral, notice that

[tex]\mathrm d(x^4+30x^2+20)=(4x^3+60x)\,\mathrm dx[/tex]

Distribute the 4 over the numerator, then substitute [tex]u=x^4+30x^2+20[/tex] and [tex]\mathrm du=(4x^3+60x)\,\mathrm dx[/tex]:

[tex]\displaystyle\frac15\int\frac{4x^3+60x}{x^4+30x^2+20}\,\mathrm dx=\frac15\int\frac{\mathrm du}u=\frac15\ln|u|+C=\frac15\ln(x^4+30x^2+20)+C[/tex]

So we have

[tex]I=\dfrac15\ln|x|+\dfrac15\ln(x^4+30x^2+20)+C[/tex]

and with some simplification,

[tex]I=\boxed{\ln\sqrt[5]{|x^5+30x^3+20x|}+C}[/tex]