intergral of (x^3 -4x + 3)/(2x) dx
would this be ln abs(x^3-4x+3) + C?
I don't really understand how to solve this problem.
d/dx (tan (x^2))
sec^2(x^2)(2x)
would this be the correct answer to find the derivative of tan (x^2)?
no, because if you differentiate your answer you don't get what you started with.
divide the 2x into each term so that your expression becomes
(1/2)x^2 - 2 + 3/(2x)
now the integral of that is
(1/6)x^3 + 2x + (3/2)lnx + c
yes, to your derivative of tan x^2
To find the integral of (x^3 - 4x + 3)/(2x), you can start by dividing each term by 2x to separate the expression. This gives us (1/2)x^2 - 2 + (3/2)x^(-1).
Next, you can integrate each term individually. The integral of x^2 is (1/3)x^3. The integral of -2 is -2x. And the integral of x^(-1) is ln|x|.
Putting it all together, the integral of (x^3 - 4x + 3)/(2x) is:
(1/3)x^3 - 2x + (3/2)ln|x| + C
Where C is the constant of integration.
Now, let's move on to the derivative of tan(x^2).
The derivative of tan(x) is sec^2(x). However, when we differentiate tan(x^2), we need to apply the chain rule.
To differentiate tan(x^2), we can think of it as tan(u) where u = x^2. Applying the chain rule states that the derivative is sec^2(u) multiplied by the derivative of u with respect to x.
Differentiating u = x^2 gives us du/dx = 2x.
Now, putting it all together, the derivative of tan(x^2) is:
sec^2(x^2) * (2x)
So your answer, sec^2(x^2)(2x), is correct for finding the derivative of tan(x^2).