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Math Help - I don't understand exactly when to use U-substitution (plus other basic questions)

  1. #1
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    I don't understand exactly when to use U-substitution (plus other basic questions)

    I understand that its the integration version of the chain-rule... but say I have

    the integral of

    (√x+1) (square root for the entire thing)

    I don't actual need to do U-sub because theres only one function, correct? So I just integrate "normally"?

    I also don't understand what the point is of putting parenthesis around a function like 2x-1 and then asking for the integral (as opposed to just asking for the integral of 2x-1)... is there any real purpose?

    Also, I get that you could solve the integral of 1/√x by moving x up and making it x to the -1/2 and integrating from there, but could you also just make it ln|√x| since the integral of 1/x is ln|x|?

    Thanks and sorry this was kind of long but these are all pretty basic questions.
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: I don't understand exactly when to use U-substitution (plus other basic questions

    Hold on.

    - For sqrt(x2+1) it is very useful to use a u-substitution, but not necessary as long as you can keep track of Chain Rule.
    - Agreed. There is no purpose for outside brackets.
    - The antiderivative of x^-1/2 is not ln|sqrt(x)|. It is 2 * x^(1/2) using Exponent Rule. (common misconception of antiderivatives involving ln)
    Thanks from skinsdomination09
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  3. #3
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    Re: I don't understand exactly when to use U-substitution (plus other basic questions

    Quote Originally Posted by MacstersUndead View Post
    Hold on.

    - For sqrt(x2+1) it is very useful to use a u-substitution, but not necessary as long as you can keep track of Chain Rule.
    - Agreed. There is no purpose for outside brackets.
    - The antiderivative of x^-1/2 is not ln|sqrt(x)|. It is 2 * x^(1/2) using Exponent Rule. (common misconception of antiderivatives involving ln)
    Thanks and one last question -

    say you're integrating any function (lets say t) and the "bounds" (what numbers you're plugging in for f(a) and f(b) to solve the integral) go from -1 to x. You don't do anything different when integrating as you would if "x" was a specified number, correct?

    so my solved integral for -1 to x for the function "t" would just be 1/2 x - 1/2. Correct?
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  4. #4
    Senior Member MacstersUndead's Avatar
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    Re: I don't understand exactly when to use U-substitution (plus other basic questions

    Correct.

    The antiderivative of t, of which you called f, is (1/2)t^2 + C. f(x) - f(-1) = (1/2)x^2 + C - (1/2) + C = (1/2)x^2 - (1/2)

    Of course, you can forget C when subtracting f(a) and f(b) since it cancels out anyway.
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    Re: I don't understand exactly when to use U-substitution (plus other basic questions

    Quote Originally Posted by MacstersUndead View Post
    Correct.

    The antiderivative of t, of which you called f, is (1/2)t^2 + C. f(x) - f(-1) = (1/2)x^2 + C - (1/2) + C = (1/2)x^2 - (1/2)

    Of course, you can forget C when subtracting f(a) and f(b) since it cancels out anyway.
    Alrighty, Can't thank you enough!!
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