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Hi all thanks for your responses.
I have gone back and rehashed over the differentiation module of calculus to get a better handle on integration, and I totally get it now. My professor told us to memorise the rules for integration by substitution/parts etc without explaining WHY or HOW the laws work, which really ticked me off because I want to understand and not memorise.
It's almost like a light bulb switched on in my head and I feel so silly for not understanding such an trivial concept of calculus!
Thank you all again
Hi all
I'm a 24 year old college student and am learning calculus over my holiday break. I studied a subject last semester which taught calculus up to partial derivatives, but it was VERY poorly taught (to the point where the university is evaluating the results of all students of the subject). I absolutely love math and I would like to study calculus and its applications in my spare time (particularly astronomy applications)
I am struggling with a few things, in particular integration by substitution. The professor showed an example of a very basic function but did not explain when this method is used, or why. From what I can gather, it is used when the function also contains its own derivative.
I had enlisted the help of a maths tutor because I hadn't studied math in about 7 years (and never any calculus).
Anyway here is my question.
I need to integrate the function f(x) = 2x/(5+x^2). Now, if we allow u(x) = 5+x^2, then du/dx = 2x. We move 2 out of the integral, but what happens to the x as the numerator? I am to understand it equals 1, as the derivative of u is removed from the integral. The integral is now 1/u which is ln (5+x^2) + C.
Now, I don't understand why exactly the derivative is gone! My maths tutor had shown me a method, but I don't understand exactly how it works. The method is as follows:
f(x)= 2x/(5+x^2) dx
u(x)= 5+x^2, du/dx= x --> du = x^2/2
=2 ∫ 1/(5+x^2) dx^2/2
=2/2 ∫ 1/(5+x^2) dx^2
= ∫ 1/(5+x^2) dx^2+5
= ∫ 1/u du
= ln|u|
=ln(5+x^2) + C
His reasoning was that we had to match the dx on the RHS with the function, and then balance it out with the coefficient outside the integral sign. I've never heard of this, and my professor or tutors at university never mentioned any such method. Actually, they haven't even explained the meaning of 'dx' on the RHS so I am really lost.
I know this is a long post but I appreciate your time and would really appreciate any help.
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