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

(√x²+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.

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)

Quote:

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Originally Posted by MacstersUndead

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)

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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 x² - 1/2. Correct?

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.

Quote:

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Originally Posted by MacstersUndead

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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Alrighty, Can't thank you enough!!