Integration Using ln

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Jason76

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POST EDITED

For instance,

\(\displaystyle \int \dfrac{x}{x^{2} + 4}dx \rightarrow\)

\(\displaystyle u = x^{2}\)

\(\displaystyle du = 2x dx\)

\(\displaystyle \dfrac{1}{2}\int \dfrac{(2)x}{u} dx\rightarrow \)

\(\displaystyle \dfrac{1}{2}\int \dfrac{1}{u} dx\rightarrow \)

\(\displaystyle \dfrac{1}{2} \int u^{-1} dx \rightarrow\) - This step and next two can be skipped, but it was written out to show why \(\displaystyle \ln\) is necessary in this situation.

\(\displaystyle \dfrac{1}{2}\dfrac{u^{-1 + 1}}{(-1 + 1)} + C\rightarrow \) - Taking the power rule leads to \(\displaystyle 0\) in the denominator which is undefined.

\(\displaystyle \dfrac{1}{2}\dfrac{u^{0}}{0} + C \rightarrow\)

\(\displaystyle \dfrac{1}{2}\ln |u| + C\rightarrow \) - Having an undefined expression in the denominator didn't cause an end to the problem, but rather was a green light to use \(\displaystyle \ln |u| + C\)

\(\displaystyle \dfrac{1}{2} \ln|x^{2} + 4| + C\)

Does this logic look ok?
 
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Jason76 said:
For instance,

\(\displaystyle \int \dfrac{x}{x^{2} + 4}dx\)

\(\displaystyle \dfrac{1}{2}\int \dfrac{2x}{u} dx\)

\(\displaystyle \dfrac{1}{2}\int \dfrac{1}{u} dx\)

\(\displaystyle \dfrac{1}{2}\dfrac{u^{-1 + 1}}{0} + C\)
Click to expand...

In the second line, where did that u come from? What is u? When you forget to define symbols that you choose, your readers need to guess.

Also, in the second line, that 2x belongs in the denominator; the numerator is x, those x's cancel, and that's where the 1/2 comes from. Don't show 1/2 and 2x together; that's confusing.

Where is your du?

In the third line, what happened to the 1/2 ? Still no du...

What is the fourth line supposed to be?

(I quit reading, at that point.)
 
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mmm4444bot said:
In the second line, where did that u come from? What is u? When you forget to define symbols that you choose, your readers need to guess.

Also, in the second line, that 2x belongs in the denominator; the numerator is x, the denominator is 2x, those x's cancel, and that's where the 1/2 comes from.

Where is your du?

In the third line, what happened to the 1/2 ? Still no du...

What is the fourth line supposed to be?

(I quit reading, at that point.)
Click to expand...

It's corrected now.
 
Jason76 said:
It's corrected now.
Click to expand...

Jason76 said:
POST EDITED

For instance,

\(\displaystyle \int \dfrac{x}{x^{2} + 4}dx \rightarrow\)

\(\displaystyle u = x^{2} \; \; \; \;\) u = x^2 + 4

\(\displaystyle du = 2x \; \; \; \;\) du = 2x dx

dx = du/(2x)

\(\displaystyle \dfrac{1}{2}\int \dfrac{2x}{u} dx\rightarrow \; \; \; \;\) Huh?
Click to expand...

Why do you put 2x in the numerator? From where is that 1/2 in front coming?

In the original integral, substitute u for x^2 + 4

Then substitute du/(2x) for dx

What do you get?
 
mmm4444bot said:
Why do you put 2x in the numerator? From where is that 1/2 in front coming?

In the original integral, substitute u for x^2 + 4

Then substitute du/(2x) for dx

What do you get?
Click to expand...

Look at again to see if written correctly. The answer is right, but perhaps it's not written out correctly.
 
Jason76 said:
\(\displaystyle \int \dfrac{x}{x^{2} + 4}dx \rightarrow\)

\(\displaystyle u = x^{2}\)

\(\displaystyle du = 2x dx\)

\(\displaystyle \dfrac{1}{2}\int \dfrac{(2)x}{u} dx\rightarrow \)
Click to expand...

Your edited post still has issues.

You seem to want u = x^2

If we do that, and substitute u for x^2, the integrand becomes:

x/(u + 4) dx

Then you somehow went to:

1/2*2x/u dx

What happened to the + 4 in the denominator?



In my previous reply, I suggested the substitution u = x^2 + 4, instead.

That sub makes the integrand x/u

:cool:

Next, sub du/(2x) for dx

That makes the integrand x/u * du/(2x) and the x's cancel

leaving 1/2 * the integral of 1/u du

Questions?
 
Jason76 said:
POST EDITED

For instance,

\(\displaystyle \int \dfrac{x}{x^{2} + 4}dx \rightarrow\)

\(\displaystyle u = x^{2}\)

\(\displaystyle du = 2x dx\)

\(\displaystyle \dfrac{1}{2}\int \dfrac{(2)x}{u} dx\rightarrow \) ..... do not do that!

Replace all the 'x' s - along with 'dx' and the integration limits (in this case the integration does not have limits) simultaneously.

\(\displaystyle du = 2\sqrt{u} \ \ dx\)

\(\displaystyle dx =\dfrac{du}{ 2\sqrt{u}}\)

\(\displaystyle \displaystyle \int \frac{x}{x^{2} + 4}dx\)

\(\displaystyle = \ \displaystyle \int \frac{\sqrt{u}}{u + 4} \ \ \frac{du}{ 2\sqrt{u}}\)

\(\displaystyle = \ \displaystyle \frac{1}{2}\int \frac{du}{u + 4}\)

and carry on...

\(\displaystyle \dfrac{1}{2}\int \dfrac{1}{u} dx\rightarrow \)

\(\displaystyle \dfrac{1}{2} \int u^{-1} dx \rightarrow\) - This step and next two can be skipped, but it was written out to show why \(\displaystyle \ln\) is necessary in this situation.

\(\displaystyle \dfrac{1}{2}\dfrac{u^{-1 + 1}}{(-1 + 1)} + C\rightarrow \) - Taking the power rule leads to \(\displaystyle 0\) in the denominator which is undefined.

\(\displaystyle \dfrac{1}{2}\dfrac{u^{0}}{0} + C \rightarrow\)

\(\displaystyle \dfrac{1}{2}\ln |u| + C\rightarrow \) - Having an undefined expression in the denominator didn't cause an end to the problem, but rather was a green light to use \(\displaystyle \ln |u| + C\)

\(\displaystyle \dfrac{1}{2} \ln|x^{2} + 4| + C\)

Does this logic look ok?
Click to expand...
.
 
Last edited by a moderator:
Or, to avoid the square roots:

If you let \(\displaystyle u = x^2, \ \ \) instead of the way that mmm4444bot showed (and that I also would have used),

then \(\displaystyle du = (2x)dx\)

\(\displaystyle \dfrac{du}{2} = (x)dx\)



\(\displaystyle \int \dfrac{x}{x^2 + 4}dx\)


\(\displaystyle = \ \int \bigg(\dfrac{1}{u + 4}\cdot \dfrac{du}{2}\bigg)\)


\(\displaystyle = \ \dfrac{1}{2}\int \dfrac{du}{u + 4} \)



and carry on . . .
 
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Here is another way that somebody showed me thru a pm. In this case, the \(\displaystyle dx\) is replaced by \(\displaystyle du\) in the numerator.

\(\displaystyle \displaystyle\int\frac{x}{x^2 + 4}\,dx\)
We have: .\(\displaystyle \displaystyle \int\frac{x\,dx}{x^2+4}\)


Your substitution is off . . .

Let \(\displaystyle u \,=\,x^2 \quad\Rightarrow\quad du \,=\,2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\frac{1}{2}du\)

Substitute: .\(\displaystyle \displaystyle \int\frac{\frac{1}{2}du}{u + 4} \;=\;\tfrac{1}{2}\int\frac{du}{u+4} \;=\; \tfrac{1}{2}\ln|u+4| +C \)

Back-substitute: .\(\displaystyle \frac{1}{2}\ln(x^2+4) +C\)
Click to expand...
 
Jason76 said:
Here is another way that somebody showed me thru a pm. In this case, the \(\displaystyle dx\) is replaced by \(\displaystyle du\) in the numerator....... Incorrect statement
Click to expand...

In all the examples - x dx is replaced by ½ du - when x2 is replaced by u [dx is replaced by du/(2√u)]
 
Jason76 said:
So the poster is wrong?
Click to expand...

No, Jason, the mistake is yours. You have misinterpreted soroban's substitutions.

In other words, it's your statement (that du replaces dx) that is wrong, not soroban's work. (Subhotosh identified this, for you.)

I do not expect you to understand. For months, you're messing around with calculus stuff that's over your head because you have yet to prepare for precalculus. Is this a wise use of your time?
 
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Jason76 said:
Here is another way that somebody showed me thru a pm. In this case, the \(\displaystyle dx\) is replaced by \(\displaystyle du\) in the numerator.
Click to expand...
[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]x/([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math])d[/FONT][FONT=MathJax_Math]x[/FONT]
We have: .[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]/(x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4)[/FONT]


Your substitution is off . . .

Let [FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]⇒[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]⇒[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/2[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]u[/FONT]

Substitute: .[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math] d[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]/(u[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/2[/FONT][FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]u/([/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Main]2 *[/FONT][FONT=MathJax_Main]ln[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]C[/FONT]

Back-substitute: .[FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/2 * [/FONT][FONT=MathJax_Main]ln[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]C[/FONT]
Click to expand...

Soroban did NOTsay what you are saying.

SoroBan said what I said → x dx = 1/2 du

NOT

replace dx by du (← this is what you said and it is incorrect)

You need to read carefully - before coming to a conclusion (right or wrong).
 
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