What is the integrant for the function e^x+1/x ?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let f(x) = e^x + 1/x

Then F(x) = intg f(x):

==> F(x) = intg ( e^x + 1/x) dx

                 = intg e^x dx  + intg 1/x  dx

We know that:

intg e^x = e^x

intg 1/x = ln x

==> F(x) = e^x + ln x + C

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The integrand is the result of differentiation of F(x), namely the function f(x).

Int f(x) dx = F(x)

or

F'(x) = f(x)

Since the integral of f(x) is e^x + 1/x, then we have to differentiate the result to determine the expression of f(x).

So, we'll compute the first derivative of the expression resulted after we've integrated f(x).

We'll note the result as F(x) =  e^x + 1/x

F'(x) = ( e^x + 1/x)'

F'(x) = (e^x)' + (1/x)'

We'll compute the first derivative of (1/x) applying the quotients rule:

(f/g)' = (f'*g - f*g')/g^2

(1/x) = (1'*x - 1*x')/x^2

(1/x) = (0 - 1)/x^2

(1/x) = -1/x^2

F'(x) = e^x - 1/x^2

But F'(x) = f(x)

So, f(x) = e^x - 1/x^2

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