1. ## Integration by Substitution.

Hey there, the given question is ..

$\int^2_1 \sqrt{x^2 + 1} \frac{dx}{x}$

I am confused with the $\frac{dx}{x}$ part.

Is the above equivalent to ..

$\int^2_1 \sqrt{x^2 + 1} \frac{1}{x} dx$ ?

If so, the approach for this question would be ..

Letting $x = sinh^2 (x)$?

-----

What is this question trying to say? Why are the boundaries in x variable while the function is in t?

$\frac{d}{dx} \int^{e^x}_{sin x} arcsin(t) log(t) dt$

Thank you!

2. For the last one, it is the second fundamental rule of calc.

Here is a way to look at it:

$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$

We get: $sin^{-1}(e^{x})log(e^{x})e^{x}-sin^{-1}(sin(x))log(sin(x))cos(x)$

Now, simplify.

For the first one, that should be $\int\frac{\sqrt{x^{2}+1}}{x}dx$.

let $x=tan(t), \;\ dx=sec^{2}(t)dt$

3. Oh, so it is true that the $\frac{dx}{x}$ is equivalent to $\frac{1}{x} dx$?

Thanks a lot for your help!!

4. one more question,

is differentiating $sin[arctan(x)]$
= $\frac{cos[arctan(x)]}{1 - x^2}$?

is there anyway to further simplify this?

5. Hello, pearlyc!

Is the derivative of $\sin[\arctan(x)] \:=\:\frac{\cos[\arctan(x)]}{1 - x^2}$ ? . . . . no

Is there anyway to further simplify this? . . . . yes
The derivative is: . $\frac{\cos[\arctan(x)]}{1\: {\color{red}+}\: x^2}$

$\text{We have: }\;\cos\underbrace{[\arctan(x)]}_{\text{angle }\theta}$

Then: . $\theta \:=\:\arctan(x) \quad\Rightarrow\quad\tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

So $\theta$ is in this right triangle:
Code:
                        *
____      *  *
√1+x²   *     *
*        * x
*           *
* θ            *
*  *  *  *  *  *  *
1

We have: . $opp = x,\;adj = 1$
From Pythagorus, we have: . $hyp = \sqrt{1+x^2}$
Hence: . $\cos\theta \:=\:\frac{1}{\sqrt{1+x^2}}$

And your answer becomes: . $\frac{\frac{1}{\sqrt{1+x^2}}}{1+x^2} \;=\;\frac{1}{(1+x^2)^{\frac{3}{2}}}$

6. Hello, pearlyc!

Oh, so it is true that the $\frac{dx}{x}$ is equivalent to $\frac{1}{x} dx$?

Sorry, but I have to snicker . . .

You're in Calculus and you're not sure if: . $\frac{y}{2} \:=\:\frac{1}{2}y$ . . . . ?

7. For the first one, that should be .

let

For that one, I got $\frac {sec^2 t}{sin t}$
Am I on the right track ?

To Soroban, it's okay to snicker It's through this I'll learn right! Tee hee. Thanks for your help too. I was out of touch for a long time, so I just wanna clear my doubts and reassure myself Thank you!