What Techniques Are Needed to Solve These Calculus Problems?

[RJLiberator]

f(x) = 1/xln(x)
g(x) = e^x sqrt(1+e^x)

Hm.

I guess, my first question is: Do I need to use U substitution?

 

[PeroK]

f(x) = 1/xln(x)
g(x) = e^x sqrt(1+e^x)

Hm.

I guess, my first question is: Do I need to use U substitution?

What substitutions are you thinking of using?

 

[RJLiberator]

Well, that's where I am kinda stuck.

Let's take the first one for example.

If I were to substitute U as ln(x) I would get:

U = ln(x)
du = 1/xdx

Working that out, I get 1/u*du

I suppose that works. Seems to check out with me.

Then it becomes ln(u) and the answer becomes ln(ln(x))

Um, wow. That is the right answer, that was a LOT easier then I made it out on my paper 0_0. It's as if I went on this site and magic happens.

OKAY, Cool, let's try example two:

e^x sqrt(1+e^x)

My two possibilities for U seem to be either e^x or sqrt(1+e^x)
Lets try U=e^x and du = e^x dx
Hm, on first impression, this does not make sense, but let's see:
du sqrt(1+u)
(2(1+u)^3/2)/3
Inputing back for e^x

(2(1+e^x)^3/2)/3

That seems to be the correct answer...

Is my reasoning correct?

 

[PeroK]

Yes, looks like you got it. Don't forget the constant of integration, though.

You may want to take a quick look at using Latex to post mathematics. E.g:

$$\int e^x \sqrt{1 + e^x}dx = \int \sqrt{1 + u}du = \frac{2}{3}(1+u)^{\frac{3}{2}} + C$$
If you "quote" my post, you'll get the Latex I used and see how it's done.

 

[RJLiberator]

PeroK, thank you kindly for all of your help here. I will look at using Latex for future posts now that I will likely be posting more here due to semester starting :).

Kind regards.

 

[RJLiberator]

Yes, looks like you got it. Don't forget the constant of integration, though.

You may want to take a quick look at using Latex to post mathematics. E.g:

$$\int e^x \sqrt{1 + e^x}dx = \int \sqrt{1 + u}du = \frac{2}{3}(1+u)^{\frac{3}{2}} + C$$
If you "quote" my post, you'll get the Latex I used and see how it's done.

Okay, one more quick question on a different question (I don't feel it is warranted to create a thread):

True or False?
For any real numbers a and b:
$\sqrt{(a+b)^2} = a+b$

My problem with this problem is that (a+b)^2 = (a+b)(a+b) = a^2+2ba+b^2 OR does it equal a^2+b^2

I am leaning towards thinking this statement is TRUE as 3+6 = 9 raised to the second power = 81. Square root of 81 = 9 and 3+6 = 9 on the other side.

Correct?

 

[PeroK]

Okay, one more quick question on a different question (I don't feel it is warranted to create a thread):

True or False?
For any real numbers a and b:
$\sqrt{(a+b)^2} = a+b$

My problem with this problem is that (a+b)^2 = (a+b)(a+b) = a^2+2ba+b^2 OR does it equal a^2+b^2

I am leaning towards thinking this statement is TRUE as 3+6 = 9 raised to the second power = 81. Square root of 81 = 9 and 3+6 = 9 on the other side.

Correct?

What about negative numbers?

 

[RJLiberator]

Oh, I am sorry Perok - I mean the absolute values of a and b on the right side, let me rewrite.

True or False?
For any real numbers a and b:
$\sqrt{(a+b)^2} = |a|+|b|$

 

[micromass]

Oh, I am sorry Perok - I mean the absolute values of a and b on the right side, let me rewrite.

True or False?
For any real numbers a and b:
$\sqrt{(a+b)^2} = |a|+|b|$

Try some examples. Try various combinations of positive and negative numbers. For example, $a$ positive and $b$ positive, $a$ negative and $b$ positive, etc.

 

[PeroK]

No. For example, a = 1, b = -1.

Taking the square root of a square quite often seems to cause trouble. Note that:

$$If \ x \ge 0, \ \ then \ \ \sqrt{x^2} = x; \ \ and \ if \ x < 0, \ then \ \sqrt{x^2} = -x$$

So, now you can apply this to (a + b).

 

[RJLiberator]

I have tried many examples now, and they all check out, so it must be true.
AH.

I get it now. (a+b)(a+b) = (a+b)^2
(a+b)^2 = a^2+b^2+2ab

Ah... silly me... thanks.

Aha.

 

[Ray Vickson]

Oh, I am sorry Perok - I mean the absolute values of a and b on the right side, let me rewrite.

True or False?
For any real numbers a and b:
$\sqrt{(a+b)^2} = |a|+|b|$

Have you tried a = 1 and b = -2?