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Leya [2.2K]
2 years ago
13

Find the integral ∫√(9+x)/(9-x)

Mathematics
1 answer:
densk [106]2 years ago
6 0

I suppose you mean

\displaystyle \int \frac{\sqrt{9+x}}{9-x} \, dx

Substitute y = √(9 + x). Solving for x gives x = y² - 9, so that 9 - x = 18 - y², and we have differential dx = 2y dy. Replacing everything in the integral gives

\displaystyle \int \frac{2y^2}{18 - y^2} \, dy

Simplify the integrand by dividing:

\dfrac{2y^2}{18 - y^2} = -2 + \dfrac{36}{18 - y^2}

\implies \displaystyle \int \left(\frac{36}{18-y^2} - 2\right) \, dy

For the first term of this new integral, we have the partial fraction expansion

\dfrac1{18 - y^2} = \dfrac1{\sqrt{72}} \left(\dfrac1{\sqrt{18}-y} + \dfrac1{\sqrt{18}+y}\right)

\implies \displaystyle \frac{36}{\sqrt{72}} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy

The rest is trivial:

\displaystyle \sqrt{18} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy

= \displaystyle \sqrt{18} \left(\ln\left|\sqrt{18}+y\right| - \ln\left|\sqrt{18}-y\right|\right) - 2y + C

= \displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+y}{\sqrt{18}-y}\right| - 2y + C

= \boxed{\displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+\sqrt{9+x}}{\sqrt{18}-\sqrt{9+x}}\right| - 2\sqrt{9+x} + C}

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Please answer quickly I will give brainliest- Kelsey has a list of possible functions. Pick one of the g(x) functions below and
Marizza181 [45]

Answer:

1. Behavior

The g(x) is large and positive when x is large and positive;

The g(x) is large and negative when x is large and negative.

2. y-intercept= -18

3. The zeros are: -3, -2 and 3

Step-by-step explanation:

1. End behavior:

To find this you have to know the leading coefficient of the variable with the highest degree, and whether the degree is even or odd.  

(x + 3) = x positive, coefficient 1

(x + 2) = x positive, coefficient 1

(x − 3)=  x positive, coefficient 1

The coefficient is 1*1*1= 1 = positive

There are 3 x, so it will be x^3= odd

The behavior for odd and positive will be:

g(x) is + ∞ when x=>+∞

g(x) is -∞ when x=>-∞

or

The g(x) is large and positive when x is large and positive;

The g(x) is large and negative when x is large and negative.

2. Y-intercept

The y-intercept is the value of y when the graph touches the y-axis. The y-axis is located at x=0, so to find the y-intercept you need to put x=0 on the graph function. Some graphs can have two y-intercepts but the graph on the question only has one. The calculation will be:

y-intercept = g(0)= (0+3)(0+2)(0-3)

y-intercept = 3*2*-3

y-intercept= -18

3. Zeros

The zeros mean the value of x that will result as g(x) as zero. At this coordinate, the graph will touch the x-axis since g(0) is located on the x-axis. That is why this coordinate also called an x-intercept. The function is expressed as the product of three things. If any of them is 0, then the result will be 0. So we have 3 zeros

x_{1}+3=0

x_{1}=-3

x_{2} +2=0

x_{2}=-2

x_{3}-3=0

x_{3}=3

The zeros are: -3, -2 and 3

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3 years ago
Solve the system of equations by substitution.<br> y = 5x - 13<br> 4x + 3y = 18<br> The solution is
Rudik [331]

Answer:

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Step-by-step explanation:

1.) 4x+3(5x-13)=18

expand into 4x+15x-39=18

18+39= 57

15x+4x= 19x

57/19= 3 so x=3

substitute 3 into y= 5x - 13

and you get 2 so y=2

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3 years ago
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Answer:

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Step-by-step explanation:

2/3 / 5/1  =  

2/3 x 1/5 = 2/15

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Answer:

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Answer:

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