# Math Help - Help with indefinite integrals and differntation involing natural logs

1. ## Help with indefinite integrals and differntation involing natural logs

Find dy/dx of : Y = ln ( x * (2x + 5)^1/3 ) / 7x^2 + 1

So to put that in words you have the function y = natural log of x times the cubed root of 2x+5 divided by 7 x squared + 1.

So far I have it broken down as follows:

I re wrote it up as ln (x) + ln (2x+5)^1/3 - ln (7x^2 + 1) using properties of logarithms.

I am having problems now with the - ln (7x^2 +1) part.

Before I get to far ahead of myself am I even headed in the right direction for what the problem is asking?

Basically I take finding dy/dx meaning say taking the function y = 3x^2.

Its derivative would be 6x (dx) and dy/dx would just be 6x.

Thats my first problem I'm working on.

I have two more that I have solved and am looking to make sure they are correct.

1) the indefinite integral of (x^4 / 3 - x^5) dx

For this one I set U = 3 - X^5 so that makes du = -4x^4.

So to balance I need to include a -4 in the numerator and subsequently add a factor of -1/4 outside of the integral. which makes it look like

-1/4 indef integral of (x^4) / (3 - x^5)

So for this one my final answer looks like -1/4 ln (3 - x^5) + C

2) the indefinite integral of sec(3x) dx

u = 3x
du = 3(dx)
du/3 = dx

= 1/3 * indef integral csc(u) du

= 1/3 * ln [ sec(u) + tan(u) ] + C

= 1/3 * ln [ sec(3x) + tan(3x) ] + C

So thats where I'm at, any help would be appreciated !!

2. $y = \ln\left(\frac{x(2x+5)^{\frac{1}{3}}}{7x^2+1}\righ t)$

$y = \ln{x} + \frac{1}{3}\ln(2x+5) - \ln(7x^2+1)$

$\frac{dy}{dx} = \frac{1}{x} + \frac{2}{3(2x+5)} - \frac{14x}{7x^2+1}$

$\int \frac{x^4}{3-x^5} \, dx$

$-\frac{1}{5} \int \frac{-5x^4}{3-x^5} \, dx$

$-\frac{1}{5} \ln|3-x^5| + C$

last one is correct, except for the csc(u) you wrote ...

3. Oh yeah it should be sec(u) which integrates into ln | sec(u) + tan(u) | + C

So for the first one I was basically down to deriving what I had re arranged and the other two were correct except for me incorrectly deriving 3-x^5. It derives into 5x^4 so to balance you need a -1/5 infront of the intergrand. which makes it -1/5 ln | 3 - x^5 | + C

Ok I figured i was very close with them and this clears it right up! Thanks!

Also in the future when I need help how do I post my problems like you have ? It would be easier to have them with the proper symbols and things like how you rewrote my examples in a more formally in math terms.

Thanks again! Lookin forward to helpin and gettin help when i need it!!

4. Originally Posted by battleman13
Oh yeah it should be sec(u) which integrates into ln | sec(u) + tan(u) | + C

So for the first one I was basically down to deriving what I had re arranged and the other two were correct except for me incorrectly deriving 3-x^5. It derives into 5x^4 so to balance you need a -1/5 infront of the intergrand. which makes it -1/5 ln | 3 - x^5 | + C

Ok I figured i was very close with them and this clears it right up! Thanks!

Also in the future when I need help how do I post my problems like you have ? It would be easier to have them with the proper symbols and things like how you rewrote my examples in a more formally in math terms.

Thanks again! Lookin forward to helpin and gettin help when i need it!!
It's done using an inbuilt LaTeX compiler.

Go to the LaTeX sub-forum and read the tutorials.