Integration: int (6x^2 + 4z + 2)/(x^3 + x^2 + x); I let u = x^2 + x + 1, du/dx = 2x + 1

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HATLEY1997

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Is this the final answer or am I missing something? Do I need to take the factor of 2 and x out at the 1st step?

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HATLEY1997 said:
Is this the final answer or am I missing something? Do I need to take the factor of 2 and x out at the 1st step?
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1708629994003.png

How do you get 1/u out of that? I don't see an actual substitution; a lot just disappears for no reason.

Instead, try not factoring the denominator at all. What happens if you take u = x^3 + x^2 + 1?
 
I’ve just done this now which makes more sense as u is now equal to the numerator bracket. Just the 2 in front of it is confusing me now!
 
HATLEY1997 said:
I’ve just done this now which makes more sense as u is now equal to the numerator bracket. Just the 2 in front of it is confusing me now!
Click to expand...
Show your new work, and we'll see what you need to do next. It looks like you know what to do with a factor like that, but maybe not.
 
[imath]u=x^3+x^2+x \implies du =3x^2+2x+1[/imath]

note …
[imath]6x^2+4x+2 = 2 \, du[/imath]
 
HATLEY1997 said:
This has helped. Am I right in taking the 2 out and putting this before the integral sign?
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Yes .... you can. In other words:

\(\displaystyle \int{C*f(x)}dx \ = \ C* \int{f(x)}dx \) ...................where C is NOT a function of (x).
 
The main problem that you are having is that you are NOT replacing each and every piece of the original integral into the new variable which is u.

Your first major error is that you did not write down the whole integral. You wrote

int (6x^2 + 4z + 2)/(x^3 + x^2 + x)​

This is wrong!! The integral is int (6x^2 + 4z + 2)/(x^3 + x^2 + x)dx

There are three pieces to your integral. They are (6x^2 + 4z + 2), (x^3 + x^2 + x) and dx
You must write ALL these expressions in terms of u, if you can. Hopefully the result will be an integral that you can easily solve. If not, then you make another u-substitution and change every expression to an equivalent expression in terms of u.

Seriously, if you don't change each and every piece of the integral to match your u-sub, you will almost always get the wrong result. Another big thing is that YOU MUST write the dx!!! Otherwise you won't have a du in your integral and it will be wrong.

It is one thing not to be able to solve an integral, but it is a whole other issue if you don't rewrite (or at least try to rewrite) every piece of the integral in terms of the substitution that you made.
 
I will do a similar one for you.
HATLEY1997 said:
I’ve just done this now which makes more sense as u is now equal to the numerator bracket. Just the 2 in front of it is confusing me now!
Click to expand...
So factor it out in front of the integral and it shouldn't confuse you anymore
 
I will start you off with a similar one. The idea is that when you take the derivative of a polynomial the degree goes down by 1 !!!

[math] \int \dfrac {(3x^2 +4x + 3)dx}{(6x^3 + 12x^2 + 18x +6)}[/math]
Since the denominator has the higher degree we will let [math]u\ =\ 6x^3 + 12x^2 + 18x +6 = 6(x^3 + 2x^2 + 3x +1) [/math] [math] now\ du =6(3x^2+ 4x + 3) dx\ or\ \dfrac{du}{ 6}= (3x^2+ 4x + 3)dx [/math]
Can you try to write out the integral in terms of u?
 
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