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$integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals$

$\log_{e} (\frac{2 + \sqrt{5}}{\sqrt{2} + 1})$
$\log_{e} (\frac{\sqrt{2} + 1}{2 + \sqrt{5}})$
$\log_{e} (\frac{2 - \sqrt{5}}{\sqrt{2} - 1})$
0

hint Hint:

We are aware that differentiation is the process of discovering a function's derivative and integration is the process of discovering a function's antiderivative. Thus, both processes are the antithesis of one another. Therefore, we can say that differentiation is the process of differentiation and integration is the reverse. The anti-differentiation is another name for the integration.
Here we have given: $integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction$ and we have to integrate it. We will use the formula to find the answer.

The correct answer is: $log subscript e invisible function application open parentheses fraction numerator 2 plus square root of 5 over denominator square root of 2 plus 1 end fraction close parentheses$

Now we have given the function as $integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction$ . Here the lower limit is 1 and upper limit is 2. We know that there are some integrals of special functions which makes problem to solve easily. We will use one of them.
We will use:
$integral subscript blank superscript blank fraction numerator d x over denominator square root of x squared plus a squared end root end fraction equals log open vertical bar x plus square root of x squared plus a squared end root close vertical bar plus C w h e r e space c space i s space c o n s tan t.$
So as per the formula, we get:
$integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals large left parenthesis log open vertical bar x plus square root of x squared plus 1 squared end root close vertical bar large right parenthesis subscript 1 superscript 2 space N o w space p u t t i n g space l i m i t s space i n space i t comma space w e space g e t colon integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log open vertical bar 2 plus square root of 2 squared plus 1 end root close vertical bar space minus space log open vertical bar 1 plus square root of 1 squared plus 1 end root close vertical bar integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log open vertical bar 2 plus square root of 4 plus 1 end root close vertical bar space minus space log open vertical bar 1 plus square root of 1 plus 1 end root close vertical bar integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log open vertical bar 2 plus square root of 5 close vertical bar space minus space log open vertical bar 1 plus square root of 2 close vertical bar N o w space w e space k n o w space t h a t space log space a space minus space log space b space equals space log space a over b comma space s o space u sin g space t h i s comma space w e space g e t colon integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log open vertical bar fraction numerator 2 plus square root of 5 over denominator 1 plus square root of 2 end fraction close vertical bar R e m o v i n g space m o d comma space w e space g e t colon integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log subscript e open parentheses fraction numerator 2 plus square root of 5 over denominator 1 plus square root of 2 end fraction close parentheses$ .

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the special case and the formula of that. The integral of the given function is $integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log subscript e open parentheses fraction numerator 2 plus square root of 5 over denominator 1 plus square root of 2 end fraction close parentheses$ .

$integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals$

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the special case and the formula of that. The integral of the given function is $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

$integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals$

Maths-General

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the special case and the formula of that. The integral of the given function is $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

General

Maths-

$integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals$

= $integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x$
>>>Integration of 1 becomes x and integration of e-x becomes -e-x.
>>> = $left parenthesis 1 minus 1 over e plus 1 right parenthesis$
= (2- $1 over e$ )

$integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals$

Maths-General

General

Maths-

$integral subscript 0 superscript pi cos invisible function application 3 x d x$

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to the final answer hence we will use trigonometric formulas. The integral of the given function is $integral subscript 0 superscript straight pi cos invisible function application 3 xdx equals 0$

$integral subscript 0 superscript pi cos invisible function application 3 x d x$

Maths-General

General

chemistry-

What is X in the nuclear reaction $scriptbase N end scriptbase presubscript 7 end presubscript presuperscript 14 end presuperscript plus subscript 1 end subscript superscript 1 end superscript H rightwards arrow scriptbase O end scriptbase presubscript 8 end presubscript presuperscript 15 end presuperscript plus X ?$

chemistry-General

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chemistry-

Which of the following pairs is / are correctly matched?

Select the correct answer using the codes given below:

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The following kinetic data are provided for a reaction between A and B:

The value of the rate constant for the above reaction is equal to

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The following data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is

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1-Chlorobutane on reaction with alcoholic potash gives

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Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives

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The potential energy diagram for a reaction R → P is given below:

$increment H to the power of 0 end exponent$ of the reaction corresponds to the energy

chemistry-General

General

chemistry-

The above transformation proceeds through

chemistry-General

General

chemistry-

A graph plotted between $l o g blank k$ vs 1/T for calculating activation energy is shown by

chemistry-General

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The correct answer is:

Now we have given the function as . Here the lower limit is 1 and upper limit is 2. We know that there are some integrals of special functions which makes problem to solve easily. We will use one of them.We will use:So as per the formula, we get:.

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The following data are for the decomposition of ammonium nitrite in aqueous solutionThe order of the reaction is

1-Chlorobutane on reaction with alcoholic potash gives

1-Chlorobutane on reaction with alcoholic potash gives

Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives

Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives

The potential energy diagram for a reaction R → P is given below: of the reaction corresponds to the energy

The potential energy diagram for a reaction R → P is given below: of the reaction corresponds to the energy

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The above transformation proceeds through

A graph plotted between vs 1/T for calculating activation energy is shown by

A graph plotted between vs 1/T for calculating activation energy is shown by

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$\log_{e} (\frac{2 + \sqrt{5}}{\sqrt{2} + 1})$
$\log_{e} (\frac{\sqrt{2} + 1}{2 + \sqrt{5}})$
$\log_{e} (\frac{2 - \sqrt{5}}{\sqrt{2} - 1})$
0

The correct answer is: $log subscript e invisible function application open parentheses fraction numerator 2 plus square root of 5 over denominator square root of 2 plus 1 end fraction close parentheses$

Which of the following pairs is / are correctly matched?

Select the correct answer using the codes given below:

Which of the following pairs is / are correctly matched?

Select the correct answer using the codes given below:

The following kinetic data are provided for a reaction between A and B:

The value of the rate constant for the above reaction is equal to

The following kinetic data are provided for a reaction between A and B:

The value of the rate constant for the above reaction is equal to

The following data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is

The following data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is

The potential energy diagram for a reaction R → P is given below:

$increment H to the power of 0 end exponent$ of the reaction corresponds to the energy

The potential energy diagram for a reaction R → P is given below:

$increment H to the power of 0 end exponent$ of the reaction corresponds to the energy

A graph plotted between $l o g blank k$ vs 1/T for calculating activation energy is shown by

A graph plotted between $l o g blank k$ vs 1/T for calculating activation energy is shown by