what is the indefinite integral of ln^2(x) for R+ -> Rcan we then verify that integral(1 to 2) { ln^2(x) dx } = 2*ln^2(e/2)?

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txmedteach | High School Teacher | (Level 3) Associate Educator

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First Part: Indefinite Integral

To do this, we'll need to integrate by parts:


`int f'(x)g(x) dx = f(x)g(x) - int f(x)g'(x) dx`

We'll let f'(x) = ln(x) and g(x) = ln(x). We end up with the following as a result:

f'(x) = ln(x), so f(x) = xln(x) - x (You can verify this by taking the derivative).

g(x) = ln(x), so g'(x) = `1/x`

Now, we can put our overall problem into the above integration by parts equation:

`int ln^2(x)dx = ln(x)(xln(x) - x) - int (xln(x) - x)*1/x dx`

`int ln^2(x) dx = xln^2(x) - xln(x) - int ln(x) - 1 dx`

Now, we need to take the antiderivative of ln(x) - 1, so we just recognize that, as when we were trying to differentiate by parts that int(ln(x)dx) will be xln(x) - x (Like when we were trying to find the f(x) for our integration by parts). We then add this to the integral of -1 to get this particular indefinite integral (don't forget your constant of integration!):

`int ln^2(x) dx = xln^2(x) - xln(x) - (xlnx - x - x + C)`

`int ln^2(x) dx = xln^2(x) - 2xln(x) + 2x - C`

Keep in mind that I can arbitrarily change the sign of C because it's just a constant of integration.

We can't simplify any more, so our indefinite integral is the following:

`xln^2(x) - 2xln(x) + 2x + C`

Evaluating the definite integral

Now, to evaluate the definite integral, we just put 2 into the above result and subtract what we get if we put 1 into the above result, seen below:

`2ln^2(2) - 2*2ln(2) + 2*2 + C - (1ln^2(1) - 2*1ln(1) + 2*1 + C)`

Recall, ln(1) = 0 because e^0 = 1. Now, we can hugely simplify by getting rid of any term containing ln(1).

`= 2ln^2(2) - 4ln(2) + 4 + C - 2 - C`

`= 2ln^2(2) - 4ln(2) + 2`

Now we distribute out a "2":

`= 2(ln^2(2) - 2ln(2) + 1)`

Now, we will add some terms. Because ln(e) = 1, we can multiply everything by ln(e) as often as we wish! So, we'll multiply the 2ln(2) term by ln(e), and we'll multiply the 1 term by ln(e) twice! This gives us the following:

`= 2(ln^2(2) - 2ln(e)ln(2) + ln^2(e))`

At this point, you might be able to see that the above equation can be factored because it is a perfect square trinomial:

`= 2(ln(e) - ln(2))^2`

We now use logarithm rules (log(a) - log(b) = log(a/b)) to get the desired result, and we're done!

`=2(ln(e/2))^2 = 2ln^2(e/2)`

 

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