Maths-
General
Easy

Question

I f integral subscript 0 superscript 4 0   d x divided by left parenthesis 2 x plus 1 right parenthesis equals l o g a t h e n a

  1. 3
  2. 9
  3. 81
  4. 40

The correct answer is: 9

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integral subscript 0 superscript pi divided by 4 end superscript   fraction numerator sin invisible function application x plus cos invisible function application x over denominator 9 plus 16 sin invisible function application 2 x end fraction d x equals

So here we used the concept of integrals 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 substitution method which makes the problem to solve easily. The integral of the given function is integral subscript 0 superscript pi divided by 4 end superscript   fraction numerator sin invisible function application x plus cos invisible function application x over denominator 9 plus 16 sin invisible function application 2 x end fraction d x equalsfraction numerator log space 3 over denominator 20 end fraction

integral subscript 0 superscript pi divided by 4 end superscript   fraction numerator sin invisible function application x plus cos invisible function application x over denominator 9 plus 16 sin invisible function application 2 x end fraction d x equals

Maths-General

So here we used the concept of integrals 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 substitution method which makes the problem to solve easily. The integral of the given function is integral subscript 0 superscript pi divided by 4 end superscript   fraction numerator sin invisible function application x plus cos invisible function application x over denominator 9 plus 16 sin invisible function application 2 x end fraction d x equalsfraction numerator log space 3 over denominator 20 end fraction

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integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x equals

So here we used the concept of integrals 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 substitution method which makes problem to solve easily. The integral of the given function is fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log left parenthesis 1 half right parenthesis

integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x equals

Maths-General

So here we used the concept of integrals 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 substitution method which makes problem to solve easily. The integral of the given function is fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log left parenthesis 1 half right parenthesis

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integral subscript 0 superscript 1   sin to the power of negative 1 end exponent 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 a final answer hence we used the formula. The integral of the given function isintegral subscript 0 superscript 1   sin to the power of negative 1 end exponent invisible function application x d x equals fraction numerator straight pi minus 2 over denominator 2 end fraction.

integral subscript 0 superscript 1   sin to the power of negative 1 end exponent 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 a final answer hence we used the formula. The integral of the given function isintegral subscript 0 superscript 1   sin to the power of negative 1 end exponent invisible function application x d x equals fraction numerator straight pi minus 2 over denominator 2 end fraction.

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integral subscript 0 superscript pi divided by 4 end superscript   x sec squared 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 formula. The integral of the given function is integral subscript 0 superscript pi divided by 4 end superscript   x sec squared invisible function application x d x equals straight pi over 4 minus log square root of 2

integral subscript 0 superscript pi divided by 4 end superscript   x sec squared 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 formula. The integral of the given function is integral subscript 0 superscript pi divided by 4 end superscript   x sec squared invisible function application x d x equals straight pi over 4 minus log square root of 2

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integral subscript negative 1 end subscript superscript 1   x e to the power of 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 formula to solve. The integral of the given function isintegral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e

integral subscript negative 1 end subscript superscript 1   x e to the power of 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 formula to solve. The integral of the given function isintegral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e

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integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis

Therefore, the integration of integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis is fraction numerator 1 over denominator log a end fraction minus fraction numerator 1 over denominator log b end fraction.

integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis

Maths-General

Therefore, the integration of integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis is fraction numerator 1 over denominator log a end fraction minus fraction numerator 1 over denominator log b end fraction.

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integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction 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 substitution method to solve. The integral of the given function is integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x equals 2 over 3 square root of 5 minus square root of 2

integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction 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 substitution method to solve. The integral of the given function is integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x equals 2 over 3 square root of 5 minus square root of 2

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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

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 1 superscript 2   fraction numerator d x over denominator square root of 1 plus x squared end root end fraction 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 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.

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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.

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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

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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)

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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

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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

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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 ?

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 ?

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 The above structural formula refers to

 The above structural formula refers to

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Which of the following compounds gives a positive iodoform test?

Which of the following compounds gives a positive iodoform test?

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Identify Z in the following series 

Identify Z in the following series 

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