Can't get this expression fully simplified

[cloudy387]

In finding the following integral [MATH]\int x\sqrt[3]{5x+1}dx[/MATH] toward the end, I arrive at:

[MATH]=\frac{3(5x+1)^{4/3}}{25}\left(\frac{5x+1}{7}-\frac{1}{4}\right)+C[/MATH]
But now I need to simplify so (omitting the C constant from now on) I distribute across the bracketed portion:

[MATH]=\frac{3(5x+1)^{7/3}}{175}-\frac{3(5x+1)^{4/3}}{100}[/MATH]
And then using LCD to form a single fraction:

[MATH]=\frac{12(5x+1)^{7/3}-21(5x+1)^{4/3}}{700}[/MATH]
At this point the computer informs me that while my solution for the integral is correct, it's not fully simplified. The desired answer is (again, omitting the C constant):

[MATH]=\frac{3(5x+1)^{4/3}(20x-3)}{700}[/MATH]
I recognize that in my answer [MATH](5x+1)^{4/3}[/MATH] can be factored out but I'm not sure what to do with it. Truth be told, I don't believe my version is that much more cumbersome than the fully simplified version, but in this case that's not for me to decide I suppose.

 

[MarkFL]

Let's begin with:

[MATH]\frac{3(5x+1)^{4/3}}{25}\left(\frac{5x+1}{7}-\frac{1}{4}\right)+C[/MATH]
Now, inside the brackets let's get a common denominator and subtract:

[MATH]\frac{3(5x+1)^{4/3}}{25}\left(\frac{4(5x+1)-7}{28}\right)+C[/MATH]
Now, let's pull the 1/28 out so that the denominator on the outside becomes 700:

[MATH]\frac{3(5x+1)^{4/3}}{700}\left(4(5x+1)-7\right)+C[/MATH]
Now, within the brackets distribute:

[MATH]\frac{3(5x+1)^{4/3}}{700}\left(20x+4-7\right)+C[/MATH]
Combine like terms:

[MATH]\frac{3(5x+1)^{4/3}}{700}\left(20x-3\right)+C[/MATH]
And arrange as:

[MATH]\frac{3(5x+1)^{4/3}(20x-3)}{700}+C[/MATH]

 

[cloudy387]

Thank you! Embarrassing how long I looked at this and never thought of finding LCD inside the parentheses first! ?‍ I thought there was going to be crazy stuff with exponents going on.

 

[Harry_the_cat]

I agree that MarkFL's method is the way to go. However you can still proceed by factorising your answer as follows (in case it happens again!). I'm just going to deal with the numerator:
\(\displaystyle 12(5x+1)^{\frac{7}{3}} -21(5x+1)^{\frac{4}{3}}\)

\(\displaystyle = 3(5x+1)^{\frac{4}{3}} (4(5x+1)-7)\)

\(\displaystyle = 3(5x+1)^{\frac{4}{3}}(20x - 3)\)

 

[cloudy387]

Oh, you're right, I can factor out a 3 along with the [MATH](5x+1)^{4/3}[/MATH] which immediately clears that numerator up. Thank you, Harry_the_cat!