$integral subscript 0 superscript 1 sin to the power of negative 1 end exponent invisible function application x d x equals$

$integral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals$

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 is $integral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e$

$integral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals$

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 is $integral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e$

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

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 $fraction numerator 1 over denominator log a end fraction minus fraction numerator 1 over denominator log b end fraction$ .

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

$integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x equals$

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 $integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x equals 2 over 3 square root of 5 minus square root of 2$

$integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 2 plus 3 x end root end fraction d x equals$

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

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 $integral subscript 1 superscript 2 fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals log subscript e open parentheses fraction numerator 2 plus square root of 5 over denominator 1 plus square root of 2 end fraction close parentheses$ .

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

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

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 $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

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

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 $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

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

= $integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x$
>>>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- $1 over e$ )

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

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

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 $integral subscript 0 superscript straight pi cos invisible function application 3 xdx equals 0$

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

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

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 data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is