Integrate:
f (x)/(2x + 1) dx
let f represent integrate sign
let u = x, du = dx => dx = du
= f (u)/(2u + 1) du
= (2u + 1)^(-1) du
= (1/2)u^2 (ln|2u + 1|) + c
= (1/2)x^2 (ln|2x + 1|) + c
...what did I do wrong? The correct answer is
(1/2)x - (1/4)ln|2x + 1| + c
It seems like there was a mistake made during the integration process. Let's go through the steps again and identify the error.
Starting from the initial expression:
∫f(x)/(2x + 1) dx
We can begin by letting u = 2x + 1. This allows us to rewrite the integral in terms of u:
dx = du/2
Now, substituting u and dx back into the integral:
∫f(x)/(2x + 1) dx = ∫f(x)/(u) (du/2) = (1/2) ∫f(x)/u du
Here is where the issue arises. Instead of integrating directly with respect to u, you integrated with respect to x, treating u as a constant. This led to an incorrect result in the solution.
To carry out the integration correctly, we should consider u as a variable and integrate with respect to u. Let's continue the correct steps:
∫f(x)/(2x + 1) dx = (1/2) ∫f(u)/u du
Now, this form of the integral can be integrated using standard techniques. However, without knowing the specific form of f(x), it is challenging to provide a more precise answer.