What is the Integral of (x^5 + 1) / (x^6 + 6x) from 2 to 3?

[mathor345]

Homework Statement

Evaluate.

Integral of (x^5 + 1) / (x^6 + 6x) on x from 2 to 3.

$$\int_2^3 (x^5 + 1) / (x^6 +6x) dx$$

Homework Equations

Not sure! The book only seems to give an equation for integrals in the form of du/u, which is integral 1/u*du = ln |u| + C. It also says that if u = f(x) THEN du = f'(x)?? Is that supposed to be a logical change? All of the integrals of this type are actually du/u in the book. This problem is not and is confusing the heck out of me.

The Attempt at a Solution

It appears that if I take u = x^6 + 6x and plug in the lower and upper limit, I will get a new limit that can be utilized in the correct answer. I already know what the answer is, but I want to figure out how to get to the answer:

1/6 ln (747/76)

The upper and lower limit when plugged into u are there: ln(747/76). I don't understand where the 1/6 is coming from and I don't understand how/why those two limits appears as a fraction.

Thanks for any help!

 

[rock.freak667]

If you have u=x^6+6x then what is du? (Factor it to see what you get)

 

[mathor345]

If you have u=x^6+6x then what is du? (Factor it to see what you get)

if u is x^6 + 6x then du = 6x^5 + 6 -> 6(x^5 + 1)...

 

[SammyS]

if u is x^6 + 6x then du = 6x^5 + 6 -> 6(x^5 + 1)...

The dx is very important!

Actually, du = 6(x^5 + 1) dx

→ (1/6) du = (x^5 + 1) dx

 

[mathor345]

The dx is very important!

Actually, du = 6(x^5 + 1) dx

→ (1/6) du = (x^5 + 1) dx

Okay... I see how the 1/6 comes into the answer. I don't see why the upper and lower limit appear inside of the ln (u) though? I've never been this confused over a single problem before.

 

[Mark44]

You have to undo the substitution. You can't just substitute 2 and 3 in for u; these are values of x.

 

[mathor345]

You have to undo the substitution. You can't just substitute 2 and 3 in for u; these are values of x.

I know, I meant that those limits are substituted for x in u (f(x)).

Ok, so apparently for a definite integral

$$\int_2^3\frac{x^5 + 1}{x^6+6x}dx$$

must be converted into the form:

$$\int_2^3\frac{1}{u}du$$

So,

$$u = x^6 + 6x$$ so,

$$du = 6x^5 + 6$$

factor out $$du = 6(x^5 + 1)$$ --> $$\frac{1}{6}du$$

Plugging in the limits, x, into u:

$$u(2) = 76$$ and $$u(3) = 747$$

So,

$$\int_x^y\frac{1}{6}\frac{du}{u} = \frac{1}{6} ln \|u}|]\right_x^y$$

(x and y are 76 and 747 respectively. LaTeX is acting funny)

-> ln |747| - ln |76| -> 1/6 ln (747/76)

So my questions are:

1) Can someone explain why the 1/6 can be brought out in front like that?
2) In the last formula du/u = ln |u| why are the terms SUBTRACTED as above so allowing for the quotient rule to be used?

 

[Mark44]

I know, I meant that those limits are substituted for x in u (f(x)).

Ok, so apparently for a definite integral

$$\int_2^3\frac{x^5 + 1}{x^6+6x}dx$$

must be converted into the form:

It's not that it "must be converted" the form below; that's what you get when you do the substution u = x⁶ + 6x, and du = 6(x⁵ + 1)dx.

Notice that in the numerator above you don't quite have du, so you multiply by 6 and divide by 6 (i.e., multiply by 1) to get what you need. The 6 in the numerator goes to make up what is needed for du. The 6 in the denominator goes outside the integral as a factor of 1/6. I think you have been missing that throughout this thread.

$$\int_2^3\frac{1}{u}du$$

Not quite. As explained above, there is a factor of 1/6, so the integral is
$$\frac{1}{6}\int_{x = 2}^3\frac{du}{u}$$

So,

$$u = x^6 + 6x$$ so,

$$du = 6x^5 + 6$$

Please read the responses more carefully. This is du = (6x⁵ + 6)dx, or du = 6(x⁵ + 1))dx. If you consistently leave off the dx you will be opening yourself up to grievous problems up ahead. That's guaranteed.

factor out $$du = 6(x^5 + 1)$$ --> $$\frac{1}{6}du$$

?
du = 6(x⁵ + 1)dx ==> du/6 = (x⁵ + 1)dx

Plugging in the limits, x, into u:

No, no, not yet! You forgot to integrate!

$$u(2) = 76$$ and $$u(3) = 747$$

So,

$$\int_x^y\frac{1}{6}\frac{du}{u} = \frac{1}{6} ln \|u}|]\right_x^y$$

(x and y are 76 and 747 respectively. LaTeX is acting funny)

-> ln |747| - ln |76| -> 1/6 ln (747/76)

So my questions are:

1) Can someone explain why the 1/6 can be brought out in front like that?
2) In the last formula du/u = ln |u| why are the terms SUBTRACTED as above so allowing for the quotient rule to be used?

1) Explained above and elsewhere in this thread.
2) No, the formula isn't du/u = ln |u|; it's
$$(1/6)\int \frac{du}{u} = (1/6) ln |u| + C$$
In this case, after undoing the substitution, the right side above becomes
(1/6) ln|x⁶ + 6x|, which you evaluate at x = 3 and x = 2. (For a definite integral you don't need to include the constant C.)

This gives you (1/6){ ln(747) - ln(76)}. The two log terms can be combined using the properties of logs.

 

[mathor345]

Thanks for responses, especially Mark for the detailed response.

It's been a year since my last math class, so I was obviously missing some of the basics. I was completely ignoring the dx as if it had no meaning, and was a bit confused with moving constants in front of the integral sign. I understand what I was doing wrong. Thanks again.