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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

  1. 2 over 3 left parenthesis square root of 5 minus square root of 2 right parenthesis
  2. 2 over 3 left parenthesis square root of 5 plus square root of 2 right parenthesis
  3. 3 over 5 left parenthesis square root of 5 minus square root of 2 right parenthesis
  4. 2 over 3 left parenthesis square root of 3 minus square root of 2 right parenthesis

hintHint:

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 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d xand we have to integrate it. We will use the substitution method to find the answer.

The correct answer is: 2 over 3 left parenthesis square root of 5 minus square root of 2 right parenthesis


    Now we have given the function as integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x. Here the lower limit is 0 and upper limit is 1. We know that there is a substitution method that makes problems to solve easily. We will use that
    We have:
    integral subscript 0 superscript 1   fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x
N o w space w e space w i l l space s u b s t i t u t e space
u space equals space 3 space x space plus space 2
D i f f e r e n t i a t i n g space i t space w i t h space r e s p e c t space t o space x comma space w e space g e t colon
fraction numerator d u over denominator d x end fraction equals 3
d x equals fraction numerator d u over denominator 3 end fraction
R e space s u b s t i t u t i n g space t h i s comma space w e space g e t colon
1 third integral fraction numerator 1 over denominator square root of u end fraction d u
N o w space i n t e g r a t i n g space i t space m a n u a l l y comma space w e space g e t colon
1 third integral fraction numerator 1 over denominator square root of u end fraction d u equals fraction numerator 2 square root of u over denominator 3 end fraction
N o w space s u b s t i t u t i n g space t h e space v a l u e space o f space u comma space w e space g e t colon
fraction numerator 2 square root of 2 plus 3 x end root over denominator 3 end fraction
P u t t i n g space u p space t h e space l i m i t s comma space w e space g e t colon
fraction numerator 2 square root of 2 plus 3 left parenthesis 1 right parenthesis end root over denominator 3 end fraction minus fraction numerator 2 square root of 2 plus 3 left parenthesis 0 right parenthesis end root over denominator 3 end fraction
fraction numerator 2 square root of 5 over denominator 3 end fraction minus fraction numerator 2 square root of 2 over denominator 3 end fraction
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

    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.

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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 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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    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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    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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    >>>              = left parenthesis 1 minus 1 over e plus 1 right parenthesis
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