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marta [7]
4 years ago
13

Find the indefinite integral of

2%7D%2Bx%5E%5Cfrac%7B3%7D%7B2%7D%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limits {\frac{5}{x^\frac{1}{2}+x^\frac{3}{2}} \, dx " alt=" \int\limits {\frac{5}{x^\frac{1}{2}+x^\frac{3}{2}} \, dx " align="absmiddle" class="latex-formula">
I have been able to simplify it to \int\limits {\frac{5\sqrt{x}}{x^3+x}} \, dx but that is confusing,

I then did u-subsitution where u=\sqrt{x} to obtain \int\limits {\frac{5u}{u^6+u^2}} \, dx which simplified to \int\limits {\frac{5}{u^5+u}} \, dx, a much nicer looking integrand
however, I am still stuck

ples help
show all work or be reported
Mathematics
1 answer:
Cloud [144]4 years ago
7 0
The easiest way to calculate this integral is substitution.

$\int\dfrac{5}{x^\frac{1}{2}+x^\frac{3}{2}}\,dx=5\int\dfrac{1}{x^\frac{1}{2}+(x^\frac{1}{2})^3}\,dx=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=(\star)

Now we can substitute u=\sqrt{x} and then:

du=\dfrac{1}{2\sqrt{x}}\,dx\qquad\implies\qquad dx=2\sqrt{x}\,du=2u\,du

So:

$(\star)=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=5\int\dfrac{2u}{u+u^3}\,du=10\int\dfrac{1}{1+u^2}\,dx=

=10\arctan(u)+C=\boxed{10\arctan(\sqrt{x})+C}

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4 red balls 4 white balls what is the probability of drawing at least 1 red ball if two balls are drawn
vodomira [7]
The probability of drawing exactly one red ball is given by:
P(1\ red\ ball)=\frac{4C1\times4C1}{8C2}=\frac{16}{28}
the probability of drawing two red balls is given by:
P(2\ red\ balls)=\frac{4C2\times4C0}{8C2}=\frac{6}{28}
The probability of drawing at least one red ball is:
P(1 red ball) + P(2 red balls) = 16/28 + 6/28 = 11/14.
 The answer is \ \frac{11}{14}

7 0
3 years ago
What is the slope of this line?<br><br> - 2/3<br><br> 2/3<br><br> 1/3<br><br> - 1/3
Paul [167]

Answer:

2/3

Step-by-step explanation:

slope =(y1-y2)/(x1-x2)

slope=(1-(-1))/(6-3)

slope =2/3

7 0
3 years ago
Using 70 POINTS NEED YOUR HELP NOT SPAMMERS. To make an ice-cream cone, employees at an ice-cream shop completely fill the cone
san4es73 [151]

<u>Solution for Q1:</u>

Given is the height of ice-cream cone, h = 6 inches.

Given is the diameter of ice-cream cone, d = 2 inches. Then radius of cone would be half of diameter, r = d/2 = 1 inch.

The ice-cream without scoop would be in the shape of cone only. So we need to find the volume of cone.

We know the formula for volume of cone is given as follows :-

Volume = \frac{1}{3} \pi r^{2} h \\\\Volume = \frac{1}{3} \pi (1)^{2} (6) \\\\Volume = \frac{6\pi}{3} = 2\pi = 2(3.14) = 6.28 \;cubic \;inches.

So the volume is 6.28 ≈ 6 in³.

Hence, option A is correct i.e. 6 cubic inches.


<u>Solution for Q2:</u>

A candy is in shape of sphere with diameter, d = 0.25 inches.

Then radius of sphere, r = d/2 = 0.250/2 = 0.125 inches.

The formula for volume of sphere is given as follows :-

Volume = \frac{4}{3} \pi r^{3} \\\\Volume = \frac{4}{3} \pi (0.125)^{3} \\\\Volume = 0.00818123 \;cubic \;inches.

So, volume of one candy = 0.00818123 cubic inches.

There are total 120 candies in the machine.

Total volume of all 120 candies = 120 x 0.00818123 = 0.981748 cubic inches.

Hence, the total volume of all candies in the machine is about 0.982 in³.

8 0
3 years ago
Read 2 more answers
Complete the following proof.
Ivan

Using the circle theorems, we have proven that m ∠RTW = 15°

<h3>Circle theorems </h3>

From the question, we are to prove that m ∠RTW = 15°

In the given diagram,

measure of arc ST = 30°

∴ m ∠SRT = 30°

m ∠SRT = ∠T + ∠W ( <em>Exterior angle of a triangle equals the sum of the two remote angles</em>)

Also,

∠T = ∠W (<em>Radii of the same circle</em>)

∴ m ∠SRT = ∠T + ∠T

m ∠SRT = 2 × ∠T

30° = 2 × ∠T

∠T = 30° /2

∠T = 15°

∴ m ∠RTW = 15°  

Hence, we have proven that m ∠RTW = 15°

Learn more on Circle theorems here: brainly.com/question/27111486

#SPJ1

7 0
2 years ago
50 points to whoever answers!!
Phantasy [73]

Answer:

create a line between (0,0) and (1,3)
the unit rate is 3/2 or 1.5

Step-by-step explanation:

pls mark

5 0
1 year ago
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