Two Integrals

Hibijibi · #1 $Old$ May 11th, 2008, 06:09 PM

Not interested in the solutions, but more of how do you know what u should be and such. Trying to grasp how I should look at a problem and piece together what I should do with it.

First integral is (x-1)/(x+1) dx

And second one is xe^(-x^2) dx

Appreciate the help!

galactus · #2 $Old$ May 11th, 2008, 06:20 PM

For the first one, divide first:

$\frac{x-1}{x+1}=1-\frac{2}{x+1}$

Now, it's easy enough to integrate without subbing.

$\int{dx}-2\int\frac{1}{x+1}dx$

Hibijibi · #3 $Old$ May 11th, 2008, 06:23 PM

How did you know to divide there? Just by practice? Other problems similar to those I've seen them split up and then several u's are used. I'm assuming you did that since the top and bottom powers were equal??

TheEmptySet · #4 $Old$ May 11th, 2008, 06:32 PM

Quote:

Originally Posted by Hibijibi $View Post$

How did you know to divide there? Just by practice? Other problems similar to those I've seen them split up and then several u's are used. I'm assuming you did that since the top and bottom powers were equal??

As a general rule if the degree of the numerator is greater than or equal to the denominator divide.

I also want to point out this trick (I forget it alot), but when they have the same degree like yours this can be done.

$\frac{x-1}{x+1}=\frac{(x+1)-2}{x+1}=\frac{x+1}{x+1}+\frac{-2}{x+1}=1-\frac{2}{x+1}$

Hibijibi · #5 $Old$ May 11th, 2008, 06:34 PM

Wow very useful trick! Taking the derivative of 1 - that is much more cleaner and easier. Thanks!!! Now just some advice on the second one. Integrals are my biggest weakness for this test.

TheEmptySet · #6 $Old$ May 11th, 2008, 06:39 PM

Quote:

Originally Posted by Hibijibi $View Post$

Wow very useful trick! Taking the derivative of 1 - that is much more cleaner and easier. Thanks!!! Now just some advice on the second one. Integrals are my biggest weakness for this test.

since we know that the integral $\int e^udu=e^u+c$

What do you think would be a good sub?

Remember if you are not sure what to pick, pick something that when replaced by u will make the integal easier. If it doesn't work try something else.

Hibijibi · #7 $Old$ May 11th, 2008, 06:42 PM

Would making u= -x^2? Actually yes that would work. Because the derivative would be du = -2x dx, and we have an x. And then I counter balance the -2 with a -1/2 on the outside?

TheEmptySet · #8 $Old$ May 11th, 2008, 06:53 PM

Quote:

Originally Posted by Hibijibi $View Post$

Would making u= -x^2? Actually yes that would work. Because the derivative would be du = -2x dx, and we have an x. And then I counter balance the -2 with a -1/2 on the outside?

Yes that is correct.

Hibijibi · #9 $Old$ May 11th, 2008, 06:54 PM

Fantastic, thank you! I also have an issue in my other thread your helping me with. You've been extraordinarily helpful to me, thank you sir.

angel.white · #10 $Old$ May 11th, 2008, 07:49 PM

Quote:

Originally Posted by Hibijibi $View Post$

And second one is xe^(-x^2) dx

Practice just looking at these ones and being able to tell the answer. First thing you should see is that x is the correct power of a derivative of x^2. And second you should see that e^(whatever) will differentiate to e^(whatever) so

e^(-x^2) differentiates to e^(-x^2) and then we apply the chain rule and get -2xe^(-x^2)

And when you can see this easily, all you need to do is realize that if you multiply this by -1/2 then you will have your integral.

ie: (-1/2) *(-2xe^(-x^2)) = xe^(-x^2)

and since -1/2 is a constant, it will carry all the way through from a f(x) to f'(x)

so we simply change e^(-x^2) into (-1/2)e^(-x^2)

and now when you differentiate it, you will get the formula you are trying to integrate.

meaning $\frac d{dx} \left(-\frac 12 ~e^{-x^2}\right) = x~e^{-x^2}$

integrate both sides to get:
$-\frac 12 ~e^{-x^2} = \int x~e^{-x^2} dx$

Here are some more, try doing them without substitution (look at it to find the answer):

1. $\int 2 ~cos(x) ~e^{sin(x)}~dx$

2. $\int \frac {arctan(x)}{x^2+1}~dx$

hint for #2: $\frac d{dx} arctan(x) = \frac 1{x^2+1}$

3. $\int \frac {arctan^2(x)}{x^2+1}~dx$

4. $\int \frac {1}{(x^2+1)~arctan(x)}~dx$

hint for #4: $\frac d{dx} ln(x) = \frac 1x$ this hint is a little abstract, if you can't figure out how to apply the thought to the problem, try integrating with substitution and then differentiate your answer to see the process it went through to get to the final solution.

5. $\int x^3~e^{x^4}~{e^{e^{x^4}}}~dx$

Hibijibi · #11 $Old$ May 11th, 2008, 07:51 PM

Wow great practice problems. Just what I need! I'll work on those and post back with what I got. Thank you!

Hibijibi · #12 $Old$ May 11th, 2008, 08:19 PM

Quote:

Originally Posted by angel.white $View Post$

Practice just looking at these ones and being able to tell the answer. First thing you should see is that x is the correct power of a derivative of x^2. And second you should see that e^(whatever) will differentiate to e^(whatever) so

e^(-x^2) differentiates to e^(-x^2) and then we apply the chain rule and get -2xe^(-x^2)

And when you can see this easily, all you need to do is realize that if you multiply this by -1/2 then you will have your integral.

ie: (-1/2) *(-2xe^(-x^2)) = xe^(-x^2)

and since -1/2 is a constant, it will carry all the way through from a f(x) to f'(x)

so we simply change e^(-x^2) into (-1/2)e^(-x^2)

and now when you differentiate it, you will get the formula you are trying to integrate.

meaning $\frac d{dx} \left(-\frac 12 ~e^{-x^2}\right) = x~e^{-x^2}$

integrate both sides to get:
$-\frac 12 ~e^{-x^2} = \int x~e^{-x^2} dx$

Here are some more, try doing them without substitution (look at it to find the answer):

1. $\int 2 ~cos(x) ~e^{sin(x)}~dx$

2. $\int \frac {arctan(x)}{x^2+1}~dx$

hint for #2: $\frac d{dx} arctan(x) = \frac 1{x^2+1}$

3. $\int \frac {arctan^2(x)}{x^2+1}~dx$

4. $\int \frac {1}{(x^2+1)~arctan(x)}~dx$

hint for #4: $\frac d{dx} ln(x) = \frac 1x$ this hint is a little abstract, if you can't figure out how to apply the thought to the problem, try integrating with substitution and then differentiate your answer to see the process it went through to get to the final solution.

5. $\int x^3~e^{x^4}~{e^{e^{x^4}}}~dx$

You're telling me you can solve all of those with out any u substitution? Mind giving me a bit more to work off of? I'm very weak in the integral area. :x

Mathstud28 · #13 $Old$ May 11th, 2008, 08:24 PM

Quote:

Originally Posted by Hibijibi $View Post$

You're telling me you can solve all of those with out any u substitution? Mind giving me a bit more to work off of? I'm very weak in the integral area. :x

Ok...I will do one and if you want any more in particular done just say so

$\int\frac{dx}{(x^2+1)arctan(x)}$

all of this is u-sub but you do it in your head...if you notice that this is

$\int\frac{dx}{(x^2+1)arctan(x)}dx=\int\bigg(\ln(arctan(x))\bigg)'dx=\ln(arctan(x))+C$

Thats how all of these are done....you could have done it u-sub

Let $u=arctan(x)\Rightarrow{du=\frac{1}{1+x^2}}$

This gives us $\int\frac{du}{u}=\ln(u)+C$

back subbing we get $\ln(arctan(x))+C$

Hibijibi · #14 $Old$ May 11th, 2008, 08:39 PM

Ahh, I understand. Could you expound on 1 and 3 please?

Mathstud28 · #15 $Old$ May 11th, 2008, 08:45 PM

Quote:

Originally Posted by Hibijibi $View Post$

Ahh, I understand. Could you expound on 1 and 3 please?

Since $\frac{D[e^{\sin(x)}]}{dx}=\cos(x)e^{\sin(x)}$

We see that $2\int\cos(x)e^{\sin(x)}dx=2\int\bigg(e^{\sin(x)}\bigg)'dx=2e^{\sin(x)}+C$

and for 3

since $\frac{D[\arctan^3(x)]}{dx}=\frac{3arctan^2(x)}{1+x^2}$

We see that $\int\frac{arctan^2(x)}{1+x^2}dx=\frac{1}{3}\int\bigg(arctan^3(x)\bigg)'dx=\frac{1}{3}arctan^3(x)+C$