The Magic of Logs: Evaluating the Integral of du/u with Natural Logarithms

Int du/u or int u^-1

To evaluate the integral ∫ du/u, we can use the natural logarithm function

To evaluate the integral ∫ du/u, we can use the natural logarithm function. Remember that the integral of a function is the antiderivative of that function.

Let’s rewrite the integral in a more familiar form: ∫ du/u = ∫ 1/u du.

To solve this integral, we can use the fact that the integral of a constant multiplied by a function is equal to that constant multiplied by the integral of the function. Therefore, we can rewrite 1/u as u^(-1).

∫ 1/u du = ∫ u^(-1) du.

Now, we need to find the antiderivative of u^(-1). The antiderivative is essentially the reverse process of differentiation. In this case, the derivative of u^(-1) is -u^(-2). Therefore, the antiderivative of u^(-1) is -u^(-2)/(-2) = 1/2u.

So, the integral of 1/u is equal to (1/2u) + C, where C is the constant of integration.

Therefore, ∫ du/u = ln|u| + C.

By replacing u with the original variable (e.g., x) if necessary, the final answer will be ln|x| + C.

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