Question

$\displaystyle\sf\int\limits_0^1tan^{-1} \left\lgroup\dfrac{1}{x^2-x-1}\right\rgroup dx$


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Answer 1

$\displaystyle\sf\int\limits_0^1tan^{-1} \left\lgroup\dfrac{1}{x^2-x-1}\right\rgroup dx$

$\frac{\pi}{2} - { log }^{2} \: is \: the \: correct \: answer \: for \: \\ this \: question$

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The definite integral of a function is closely related to the antiderivative and indefinite integral of a function. The primary difference is that the indefinite integral, if it exists, is a real number value, while the latter two represent an infinite number of functions that differ only by a constant. The relationship between these concepts is discussed in the section on the Fundamental Theorem of Calculus, and you will see that the definite integral will have applications to many problems in calculus.

Answer 2

Step-by-step explanation:

write a program to input the sales made by salesman (S). lf S>=10000 , he earns the commission of 30% on the sales made. lf S>=5000 commission is 20% else commission is 10% calculate the commission in each case .

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