[tex] \sf \: please \: answer \: and \: do \: not \: spam[/tex]
¢αяσυѕєℓ :)

It's based on integrals....​

[tex] \sf \: Please \: Answer \: And \: Do \: Not \: Spam[/tex] :)It's Based On Integrals....

Answers

Answer 1

Hello Carousel!

[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]

Solve the integral.

[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]

[tex] \tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex]

First, let's take I as ⇨ [tex]\tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex].

[tex]\tt \:I = \int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x +1}} \: dx \\\tt \: I =\int \: \frac{ x + 2 }{( {x}^{2} + 2x + 1 + x + 2) \sqrt{x + 1}} \: dx \\ \tt \:I =\int \: \frac{ x + 2 }{( ({x + 1}^{2}) + x + 2) \sqrt{x + 1}}[/tex]

Let, x + 1 = m² => dx = 2mdm.

[tex]\tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: m} 2mdm \\ \tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: \bcancel{ m}} 2 \bcancel{m}dm \\ \tt \: I = \: \int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 } 2dm[/tex]

Now, divide the numerator & denominator by m²....we'll get it as...

[tex]\tt \:I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{ {m}^{2} + 1 + \frac{1}{ {m}^{2} } } \: dm \\\tt \: I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{( {m}^{2} + \frac{1}{ {m}^{2} } - 2) + 3} \: dm \\\tt \: I =2\int \: \frac{ (1 + \frac{1}{ {m}^{2} }) \: dm }{ ({m} - \frac{1}{ m } ) ^{2} + 3} [/tex]

Now, let m - 1/m be t => (1 + 1/m²) dm = dt

[tex]\tt \:I =2\int \: \frac{ dt}{ {t}^{2} + 3 } \\ \tt \:I = 2\int \: \frac{ dt}{ {t}^{2} + ( \sqrt{3}) ^{2} }[/tex]

We know, [tex]\tt \:\int \: \frac{dx}{ {x}^{2} + a ^{2} } = \frac{1}{a} tan ^{ - 1} (\frac{x}{a} ) + c[/tex]...therefore...

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{t}{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 1}[/tex]

Now, substitute the value of 't' in eq. 1..we'll get..

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{m - \frac{1}{m} }{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 2}[/tex]

Now, substitute the value of 'm' in eq. 2...we'll get...

[tex] \tt \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{ \sqrt{x + 1} - \frac{1}{ \sqrt{x - 1} } }{ \sqrt{3} } ) + c \\ \tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x + 1 - 1}{ \sqrt{3} \sqrt{x - 1} } ) + c \\ \boxed{\boxed{ \bf \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x }{ \sqrt{3 (x - 1} )} ) + c }}[/tex]

The correct answer is Option B.

__________________

The question is really long & tricky but once you get the hang of it you'll be good. Good luck!

__________________

Hope it'll help you!

ℓu¢αzz ッ


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________

x = -1, is the answer.

Hope this helps! Let me know if you have any other questions related to this problem. :)

Answer:

x= -1

Step-by-step explanation:

-6+x = -7

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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1 no

2 I don't know

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Step-by-step explanation:

1 no because the y overlaps

2 not sure

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4 it never overlaps

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Answers

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Answers

Step-by-step explanation:

[tex] { - 100}^{1} [/tex]

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer:

Q'(-4,0), R'(-4,1), S'(0,0), T'(-3,-3)

Step-by-step explanation:

To solve this, what we need to do is find out what the translation is. So if we are translating those coordinates up two units, then that means we are basically adding two to the y value.

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Answer:

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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A,C,E

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3/8×4/5

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...........

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3/10

Step-by-step explanation:

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Answers

Answer:

C. 1.2009

Step-by-step explanation:

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Step-by-step explanation:

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