# Definite integral involving the natural log function

#### Bmanmcfly

##### Member
I figured I would just add this new problem over here, rather than starting a new thread.

Im looking to solve integration leading to arctan or arcsin results.

$$\displaystyle \int_{1}^{e}\frac{3dx}{x(1+\ln(x)^2})$$

Looking at this, it feels like this has an arctan in the result, but I would have to multiply the x with the $$\displaystyle (1+\ln(x)^2)$$ and then would figure out a difference of squares to figure out the integral...

that ln(x) is what is screwing me up...

So, I guess I'm asking how working with the ln(x)^2 would be different from working with just an x.

Side question, what is the difference really between $$\displaystyle tan^{-1}, Cot and arctan$$ it seems to me that these are just different ways of saying cos/sin. Is there something else significant that I'm ignorant about?

#### MarkFL

Staff member
We actually prefer that you begin a new topic for a new question rather than tag a new question onto an existing topic, as per rule #8. I have split the topic so that your new question has its own topic.

For this definite integral, I recommend using the substitution:

$$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

Now, being mindful to change the limits in accordance with the substitution, how may you now rewrite the integral?

#### Bmanmcfly

##### Member
We actually prefer that you begin a new topic for a new question rather than tag a new question onto an existing topic, as per rule #8. I have split the topic so that your new question has its own topic.

For this definite integral, I recommend using the substitution:

$$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

Now, being mindful to change the limits in accordance with the substitution, how may you now rewrite the integral?

Woops, I was thinking it would keep things more streamline.... Not important.

As for the integral,

Thanks for the pointer, the limits would change to ln(e)- ln(1) or 1-0.
let me know if I got you right...
$$\displaystyle 3\int_{\ln(1)}^{\ln(e)}\frac{(\frac{1}{u}du)}{(1+u^2)}$$

This gives $$\displaystyle 3\tan^{-1}(u)|_{\ln(1)}^{\ln(e)}$$

Thanks again so much, you don't even know.

#### MarkFL

Staff member
You've got the right idea...I would think of the integral as:

$$\displaystyle \int_1^e\frac{3}{x(1+\ln^2(x))}\,dx=3\int_1^e \frac{1}{\ln^2(x)+1}\cdot\frac{1}{x}\,dx$$

Now, when you make the substitution, you may write:

$$\displaystyle 3\int_0^1\frac{1}{u^2+1}\,du=3\left[\tan^{-1}(u) \right]_0^1$$

#### Bmanmcfly

##### Member
You've got the right idea...I would think of the integral as:

$$\displaystyle \int_1^e\frac{3}{x(1+\ln^2(x))}\,dx=3\int_1^e \frac{1}{\ln^2(x)+1}\cdot\frac{1}{x}\,dx$$

Now, when you make the substitution, you may write:

$$\displaystyle 3\int_0^1\frac{1}{u^2+1}\,du=3\left[\tan^{-1}(u) \right]_0^1$$
Now, the other part of the question;

Is there any significant difference between $$\displaystyle \tan^{-1} , \cot, \and \arctan$$??

It seems that all mean the same thing? What's the reason to use one over the other? Cause to me it seems just preference.

#### MarkFL

Staff member
Well, we do have:

$$\displaystyle \arctan(x)\equiv\tan^{-1}(x)$$

However, the cotangent function is the multiplicative inverse of the tangent funtion, not the functional inverse, and this is a common misconception among students.

$$\displaystyle \cot(\theta)\equiv\frac{1}{\tan(x)}$$

whereas if:

$$\displaystyle x=\tan(\theta)$$ then $$\displaystyle \theta=\tan^{-1}(x)$$

usually defined where $$\displaystyle -\frac{\pi}{2}<\theta<\frac{\pi}{2}$$.

It can be a confusing notation, where for variables, we take:

$$\displaystyle x^{-1}=\frac{1}{x}$$

but for functions:

$$\displaystyle f^{-1}(x)\ne\frac{1}{f(x)}$$

and by definition:

$$\displaystyle f^{-1}\left(f(x) \right)=x$$