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Evaluate not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fractiondx

  1. fraction numerator 1 over denominator 4 square root of 3 end fraction l space l n invisible function application open vertical bar fraction numerator square root of x plus 1 end root minus square root of 3 over denominator square root of x plus 1 end root plus square root of 3 end fraction close vertical bar minus 1 fourth t a n to the power of negative 1 end exponent invisible function application left parenthesis square root of x plus 1 end root right parenthesis plus c    
  2. negative fraction numerator 1 over denominator 4 end fraction t a n to the power of negative 1 end exponent invisible function application left parenthesis square root of x plus 1 end root right parenthesis plus fraction numerator 1 over denominator 4 end fraction l n invisible function application open vertical bar fraction numerator square root of x plus 1 end root minus square root of 3 over denominator square root of x plus 1 end root plus square root of 3 end fraction close vertical bar plus C    
  3. negative fraction numerator 1 over denominator 4 square root of 3 end fraction t a n to the power of negative 1 end exponent invisible function application left parenthesis square root of x plus 1 end root right parenthesis plus fraction numerator 1 over denominator 4 end fraction l n invisible function application open vertical bar fraction numerator square root of x plus 1 end root minus 1 over denominator square root of x plus 1 end root plus 1 end fraction close vertical bar plus C    
  4. None of these    

The correct answer is: negative fraction numerator 1 over denominator 4 square root of 3 end fraction t a n to the power of negative 1 end exponent invisible function application left parenthesis square root of x plus 1 end root right parenthesis plus fraction numerator 1 over denominator 4 end fraction l n invisible function application open vertical bar fraction numerator square root of x plus 1 end root minus 1 over denominator square root of x plus 1 end root plus 1 end fraction close vertical bar plus C

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If f left parenthesis x right parenthesis equals t a n to the power of negative 1 end exponent invisible function application x plus l space l n invisible function application square root of 1 plus x end root minus l n square root of 1 minus x end root. The integral of 1 divided by 2 space f apostrophe left parenthesis x right parenthesis with respect to x to the power of 4 end exponent is -

If f left parenthesis x right parenthesis equals t a n to the power of negative 1 end exponent invisible function application x plus l space l n invisible function application square root of 1 plus x end root minus l n square root of 1 minus x end root. The integral of 1 divided by 2 space f apostrophe left parenthesis x right parenthesis with respect to x to the power of 4 end exponent is -

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If I = not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction, then I equals:

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The indefinite integral of left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent is, for any arbitrary constant -

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Let x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N, then integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x is equal to:

Let x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N, then integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x is equal to:

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Statement II : sin x and {x} are both periodic with period 2 pi and 1 respectively.

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Statement 1 : f : R rightwards arrow R and f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponentis bijective.
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Statement 1 : f : R rightwards arrow R and f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponentis bijective.
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maths-General
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Assertion (A) : Graph of open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text  and  end text x comma y element of R close curly brackets
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Assertion (A) : Graph of open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text  and  end text x comma y element of R close curly brackets
Reason (R) : In the expression am/n, where a, m, n × J+, m represents the power to which a is be raised, whereas n determines the root to be taken; these two processes may be administered in either order with the same result.

Maths-General
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maths-

Assertion: The period of f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket is 1/2.
Reason: The period of x – [x] is 1.

Assertion: The period of f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket is 1/2.
Reason: The period of x – [x] is 1.

maths-General
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Assertion : Fundamental period of c o s invisible function application x plus c o t invisible function application x text  is  end text 2 pi.
Reason : If the period of f(x) is T subscript 1 end subscript and the period of g(x) is T subscript 2 end subscript, then the fundamental period of f(x) + g(x) is the L.C.M. of T subscript 1 end subscript and T

Assertion : Fundamental period of c o s invisible function application x plus c o t invisible function application x text  is  end text 2 pi.
Reason : If the period of f(x) is T subscript 1 end subscript and the period of g(x) is T subscript 2 end subscript, then the fundamental period of f(x) + g(x) is the L.C.M. of T subscript 1 end subscript and T

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