1. The integral of: [Square Root(x^2 - 36)] / x^2 dx


2. The integral of: (y^3 + 2y^2 - 1) / (3y^4 + 12y^2) dy


3. The integral of: (lnw)^2 dw


For Number 1 this is how far I got:
x = asectheta therefore x = 6sectheta
the sq root part = 6tantheta

Therefore:
integral: (6tantheta / 36sec^2theta) dx

1/6 integral (tantheta / sec^2theta) dx


Not sure how to go about (tantheta / sec^2theta). I would need an identity i believe and be able to apply it. Also I am not 100% sure what the deriv of 6sectheta is (i know the deriv of secx = secxtanx).

Any help is appreciated!

-Thanks for your time

If you just want the answer you can use wolfram alpha

isnt the 3rd one pretty easy? or i am mistaking it with something else

This post was edited by Rimi on Oct 12 2014 02:03am

Damn, that looks hard. Sorry I can't help.

Need help on completing #1

*I have the answer using Wolfram (posted after my updated work, and to get to the answer an identity just needs to be used):

Here is the problem:
1. The integral of: [Square Root(x^2 - 36)] / x^2 dx

x = asectheta therefore x = 6sectheta
the sq root part = 6tantheta

Therefore:
integral: (6tantheta / 36sec^2theta) dx

1/6 integral (tantheta / sec^2theta) dx

Since dx = 6SecthetaTantheta, we plug this in for dx and get:

integral (tantheta / sec^2theta) SecthetaTantheta ----> Here i believe you would convert everything to sin and cos using an identity

Answer is: Sintheta + TanthetaSectheta

If anyone can help me get to the answer from where I left off, that would be really appreciated!

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Is my strategy for Problem number 2 correct?

Here is the problem:
The integral of: (y^3 + 2y^2 - 1) / (3y^4 + 12y^2) dy

Strategy ---------> Integration of rational functions by partial fractions

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Is this work and answer correct for #3?

Not sure if this answer is correct, if anyone can verify that would be great!
Here is problem with work and answer:

3. The integral of: (lnw)^2 dw
=2ln(x) times d/dx(ln(x))
=2(1/x) times ln(x)
=2[ln(x)] / x ----> Answer

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Thanks again guys for your time!

i got this for the third one

F(w) = w*(ln(w))² - 2w*ln(w)+2w

you can shorten it though

1. I would honestly just use integration by parts where u = sqrt(x^2-36) and v' = 1/x^2 because you'll have to do a integration by parts with trig functions anyway.

2. Yeah my first thought for solving it would be turning into partial fractions

3. Im not quite sure what you did but i would also use integration by parts here again where u = ln(w^2) and v' = 1

If anyone can complete these with work, i would greatly appreciate it. I need to have this done by 3:45pm today.

If you scratch my back, i will scratch yours

This post was edited by CamelFinger on Oct 13 2014 09:51am

Nice, GL.