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storchak [24]
2 years ago
10

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

he inverse tangent of the quantity x squared divided by 4, plus C one fourth times the natural log of x to the 4th power plus 16, plus C one fourth times the inverse tangent of the quantity x squared divided by 4, plus C the product of negative one fourth and 1 over the quantity squared of x squared plus 16, plus C
Mathematics
1 answer:
Anika [276]2 years ago
4 0

Answer:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

Step-by-step explanation:

Given

\int\limits {\frac{x}{x^4 + 16}} \, dx

Required

Solve

Let

u = \frac{x^2}{4}

Differentiate

du = 2 * \frac{x^{2-1}}{4}\ dx

du = 2 * \frac{x}{4}\ dx

du = \frac{x}{2}\ dx

Make dx the subject

dx = \frac{2}{x}\ du

The given integral becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

Recall that: u = \frac{x^2}{4}

Make x^2 the subject

x^2= 4u

Square both sides

x^4= (4u)^2

x^4= 16u^2

Substitute 16u^2 for x^4 in \int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du

Simplify

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

In standard integration

\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)

So, the expression becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)

Recall that: u = \frac{x^2}{4}

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

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Nutka1998 [239]

Answer:

\frac{19}{8}

Step-by-step explanation:

First, multiply the two denominators together to get the shared denominator, which will allow you to actually subtract the two numbers. 4 × 8 = 32, so the shared denominator will be 32.

Then, cross-multiply. 11 × 8 = 88, so \frac{11}{4} becomes \frac{88}{32}. 3 × 4 = 12, so \frac{3}{8} becomes \frac{12}{32}.

Now, you can subtract.  \frac{88}{32} -\frac{12}{32} =\frac{76}{32}

Finally, simplify. \frac{76}{32} =\frac{19}{8}. The final difference is \frac{19}{8}.

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Step-by-step explanation:

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2 years ago
An automobile manufacturer is preparing a shipment of cars and trucks on a cargo ship that can carry 21,600 Tons. The cars weigh
inysia [295]

Answer:

Number of trucks (y) that can be shipped if the number of cars (x) is known=\frac{21,600-3.6x}{7.5}

Step-by-step explanation:

A cargo ship can carry 21,600 Tons.

Let x denotes number of cars and y denotes number of trucks.

The cars weigh 3.6 tons each and the trucks weigh 7.5 tons each.

Therefore,

3.6x+7.5y=21,600

Now write y in terms of x.

7.5y=21,600-3.6x\\y=\frac{21,600-3.6x}{7.5}

Number of trucks (y) that can be shipped if the number of cars (x) is known=\frac{21,600-3.6x}{7.5}

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Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ
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2x - 3y = - 13

the equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

First obtain the equation in slope-intercept form

y = mx + c ( m is the slope and c the y-intercept )

calculate m using the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (1, 5 ) and (x₂, y₂ ) = (- 2, 3 )

m = \frac{3-5}{-2-1} = \frac{-2}{-3}= \frac{2}{3}

y = \frac{2}{3} x + c ← partial equation

to find c substitute either of the 2 points into the partial equation

using (1, 5 ), then

5 = \frac{2}{3} + c ⇒ c = \frac{13}{3}

rearrange the equation into standard form

multiply through by 3

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2x - 3y = - 13 ← in standard form


3 0
2 years ago
Read 2 more answers
Please help easy stuff.......
nataly862011 [7]

To evaluate a function at a given input, you have to substitute every occurrence of the variable with that particular value.

So, for the first function, you have

T(x) = -x-2 \implies T(-4) = -(-4)-2 = 4-2 = 2

The name of the variable is obviously irrelelvant, so the same goes for the second function:

w(x) = 14-6x \implies w(-3) = 14-6\cdot(-3) = 14 + 18 = 32

3 0
2 years ago
Read 2 more answers
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