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Lelu [443]
3 years ago
7

Can anyone help me integrate :

Mathematics
1 answer:
worty [1.4K]3 years ago
3 0
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
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This is my last question please help asap!!
Ivanshal [37]

The average rate of change of function g(x)=5(2)^{x} from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

The correct option is (A).

What is the average rate of change of a function?

The average rate at which one quantity changes in relation to another's change is referred to as the average rate of change function.

Using function notation, we can define the Average Rate of Change of a function f from a to b as:

                                     rate(m) = \frac{f(b)-f(a)}{b-a}

The given function is  g(x) = 5(2)^{x},

Now calculating the average rate of change of function from x = 1 to x = 2.

                               m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(2)-g(1)}{2-1}\\\\m=\frac{5(2)^{2}-5(2)^1}{2-1}\\ m=\frac{10}{1} \\m=10

Now, calculate the average rate of change of function from x = 3 to x = 4.

                                 m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(4)-g(3)}{4-3}\\\\m=\frac{5(2)^{4}-5(2)^3}{4-3}\\ m=\frac{40}{1} \\m=40

The jump from m = 10 to m = 40 is "times 4".

So option (A) is correct.

Hence, The average rate of change of function g(x)=5(2)^{x} from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

To learn more about the average rate of change of function, visit:

brainly.com/question/24313700

#SPJ1  

7 0
1 year ago
write the equation of the line that passes through the points (-4,4) and (6,9) and figure out what y=mx+b for this equation woul
jeka57 [31]
Hello there! So, y = mx + b is in slope-intercept form, where m represents the slope, b represents the y-intercept, and y and x remain unfilled. First off, let's solve for the slope. The formula for slope is y2 - y1 / x2 - x1, where you subtract the first x and y coordinates from the second x and y coordinates. So it would be formed like this:

9 - 4 / 6 - (-4)

Let's subtract. 9 - 4 is 5. 6 - (-4) is 10. 5/10 is 1/2 in simplest form. The slope for this equation is 1/2. Now, let's find the y-intercept. We will find that by plugging one of the points into the equation and solving for b. The x and y coordinates will be filled in by that coordinate. Let's use (-4, 4) for this problem. We will also plug in the slope. In this case, the problem will look like this:

4 = (1/2)(-4) + b

Now, let's multiply 1/2 and -4 to get -2. Now, to get b by itself, subtract 2 to both sides to isolate the b. -2 + 2 cancels out. 4 + 2 is 6. b = 6. There. The equation of the line in slope-intercept form is y = 1/2x + 6.
6 0
3 years ago
Area of circle <br> sector and finding the percentage
mestny [16]

Answer:2592/25

Step-by-step explanation: Big circle is pi*r^2= (4+3+3)^2*pi =314.15 minus 4^2*pi=50.26 equals 263.89. the big circle minus the first 2 layers of the circle = the outer circle. 314.15-(4+3)^2*pi=160.21 263.89-160.21=2592/25

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3 years ago
On a coordinate plane, 2 right triangles are shown. The first triangle has points A (negative 1, 3), B (negative 1, 1), C (3, 1)
Mars2501 [29]
The second and last answer choice
7 0
2 years ago
0.61m - 1.51m = 9 Solve for x
worty [1.4K]
X=(-10 ) Hope this helps!!! :)
4 0
3 years ago
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