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

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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 **

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**