Integrate from e^2 to e of tthis funtion: 1/(x)ln(x) dx
To integrate the function 1/(xln(x)) with respect to x from e^2 to e, you can follow these steps:
1. Start by noting that the integral of a function f(x) over an interval [a, b] can be found using the fundamental theorem of calculus:
∫[a, b] f(x) dx = F(b) - F(a),
where F(x) is the antiderivative (also known as the indefinite integral) of f(x).
2. To find the antiderivative of 1/(xln(x)), we can perform a u-substitution. Let u = ln(x). Then, du/dx = 1/x, which implies that dx = du/(1/u).
Substituting these into the integral, we have:
∫ 1/(xln(x)) dx = ∫ 1/u * (du/(1/u)) = ∫ du = u + C,
where C is the constant of integration.
3. Going back to the original variable x, we have:
∫ 1/(xln(x)) dx = ln(x) + C.
4. Now, we can evaluate the definite integral from e^2 to e by plugging in the limits:
∫[e^2, e] 1/(xln(x)) dx = [ln(x)]|[e^2, e] = ln(e) - ln(e^2),
where "[ln(x)]|[a, b]" denotes evaluating ln(x) at the limits a and b.
5. Simplifying the expression:
∫[e^2, e] 1/(xln(x)) dx = ln(e) - ln(e^2) = 1 - 2ln(e) = 1 - 2 = -1.
Therefore, the definite integral of 1/(xln(x)) with respect to x from e^2 to e is -1.