Challenged by my teacher: integration by substitution

In summary, Pablo struggled with integrating x^2+1 through substitution, and was given an F if he didn't solve it tomorrow. He was told that this is what he gets for being one of the kids who always understands in class.
  • #1
Juan Pablo
40
0
I just learned how to integrate through substitution and I was challenged by my teacher with an apparently easy problem but I'm really struggling with it.
He said he will give me an F if I don't solve it for tomorrow, I guess this is what I get by being the one who always understand in class

Exposed below

Homework Statement



[tex]\int (x^2+1)^2 dx[/tex]

Homework Equations



[tex]x^2+1=u[/tex]

[tex]\frac{du}{2x}=dx[/tex]

[tex]x=\sqrt{u-1}[/tex]

The Attempt at a Solution



[tex]\int u^2(\frac{du}{2x}) = [/tex]

[tex]\int u^2(\frac{du}{2\sqrt{u-1}}) = [/tex]

[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

Now I finally integrate

[tex]
\frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]

[tex]\frac{1}{2} ln(\frac{\sqrt{u-1}}{u^2})[/tex]

[tex]\frac{1}{2} ln(\frac{x}{(x^2+1)^2})[/tex]But if I integrate it solving the binomial first I get a totally different answer and If I derive my result I don't get [tex](x^2+1)^2[/tex]

Anybody?
 
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  • #2
You can check your answer by just foiling out [itex](x^2+1)^2[/itex]. So you really shouldn't be getting natural logs...

Perhaps he's just trying to get you to use u substitution when you don't have to?
 
  • #3
Yes, that was my first thought but he said that I must use substitution, and this is the only way I think its possible.

Thanks for your answer!
 
  • #4
Juan Pablo said:
I just learned how to integrate through substitution and I was challenged by my teacher with an apparently easy problem but I'm really struggling with it.
He said he will give me an F if I don't solve it for tomorrow, I guess this is what I get by being the one who always understand in class

Exposed below

Homework Statement



[tex]\int (x^2+1)^2 dx[/tex]

Homework Equations



[tex]x^2+1=u[/tex]

[tex]\frac{du}{2x}=dx[/tex]

[tex]x=\sqrt{u-1}[/tex]

The Attempt at a Solution



[tex]\int u^2(\frac{du}{2x}) = [/tex]

[tex]\int u^2(\frac{du}{2\sqrt{u-1}}) = [/tex]

[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

Now I finally integrate
[tex]
\frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]
But, that is wrong. That is not the value of the integral.
[tex]\frac{1}{2} ln(\frac{\sqrt{u-1}}{u^2})[/tex]

[tex]\frac{1}{2} ln(\frac{x}{(x^2+1)^2})[/tex]


But if I integrate it solving the binomial first I get a totally different answer
you get a different answer because you get the right answer, which is different from the wrong answer.
 
  • #5
I'm just learning this, is there anybody out there willing to show me how to integrate it correctly through substitution?
 
  • #6
Juan Pablo said:
[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

Now I finally integrate

[tex]
\frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]
That's clearly not right.
 
  • #7
Isn't this one of the formulas?

[tex]\int \frac{dx}{x} = ln(x)[/tex]

I'm really trying, and I would appreciate answers with more than one line. I know it's wrong but if somebody would be so kind and try to explain it to me?

I just learned integration three days ago and I learned substitution this morning
 
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  • #8
Juan Pablo said:
Isn't this one of the formulas?

[tex]\int \frac{dx}{x} = ln(x)[/tex]

I'm really trying, and I would appreciate answers with more than one line. I know it's wrong but if somebody would be so kind and try to explain it to me?

I just learned integration three days ago and I learned substitution this morning

Hi Pablo,

Sorry if it seems that we're being a bit short with our answers. The homework forum isn't a place where others do your homework for you, just offer advice.

In any case, You're right that [itex]\int \frac{dx}{x} = \ln(x)[/itex] - but this isn't what you have, is it? Instead, you have [itex]\int \frac{dx}{x^{-2}\sqrt{1-x}}[/itex] - that is, [itex]\int \frac{dx}{f(x)} \ne \ln(f(x))[/tex] in general.

As for how to integrate that thing, I'm afraid I'm stumped without further substitution. Integrals.com and the back of Stewart's calculus book gives [itex]2/30 \sqrt{u-1}(3u^2+4u+8)[/itex] the answer, which, when substituting back in your original substitution gives the correct answer.

EDIT: Stewart's Calculus has the following:

[tex]\int \frac{u^2du}{\sqrt{a+bu}}=\frac{2}{15b^3}(8a^2+3b^2u^2-4abu)\sqrt{a+bu}+C[/tex].

For your integral, [itex]a=b=1[/itex]
 
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  • #9
Why would you think of trying to solve that integral by substitution? Any substitution (at least that I can think of) would just make the integrand more complicated. If I were you, I'd just expand and solve it in two lines (of course you can make the substitution u=x if your teacher demands a substitution ;) ) and move on to trying to solve more complicated problems that will help you hone your techniques for the future.
 
  • #10
Yes I know you are not doing my homework, and it's not really my homework, it's a challenge put by my teacher just to me. I'm also a permanent user in other forums about technical topics and I always try to be as detailed and kind as possible when I answer questions. Even If they are homework I give hints and tips to the poster.

Mathematica gave me the same answer, thank you however

Cristo: Yes, it's a lot easier without substitution and I can solve it that way right now, but I'm asked to use substitution. I think I'll stick to u=x like you said. Thanks
 
  • #11
Thanks for your edit and formula, I think I will use it although I haven't seen that formula on class.
 
  • #12
Juan Pablo said:
[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}}[/tex]

Juan Pablo said:
[tex]\int \frac{dx}{x} = ln(x)[/tex]

You can't apply that rule -- [itex]du / (u^{-2}\sqrt{u-1})[/itex] simply doesn't match the pattern [itex]dx / x[/itex].
 
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  • #13
My teacher seems to be a moron, he used the formula a couple of times like I did. I think this led to my confusion. I believe the correct way is with PingPong's formula
 
  • #14
Thanks Hurkyl PingPong, your help is really appreciated
 
  • #15
Juan Pablo said:
I believe the correct way is with PingPong's formula
But the point of substitutions is to make an integral simpler to compute, not make it harder! If your teacher did intend that to be the solution, then I have no idea what he is thinking-- what a waste of time!
 

1. What is integration by substitution?

Integration by substitution is a method used in calculus to find the integral of a function by replacing the variable with a new variable. This technique is also known as the u-substitution method.

2. Why is integration by substitution useful?

Integration by substitution is useful because it allows us to solve integrals that would otherwise be difficult or impossible to solve. It also simplifies the integration process by converting the integral into a form that is easier to solve.

3. How do I perform integration by substitution?

To perform integration by substitution, follow these steps:

1. Identify a part of the integrand that can be replaced with a new variable.

2. Substitute the new variable and its derivative into the integral, making sure to also change the limits of integration.

3. Solve the integral with respect to the new variable.

4. Replace the new variable with the original variable in the final answer.

4. What are some common mistakes when using integration by substitution?

Some common mistakes when using integration by substitution include:

- Choosing the wrong variable to substitute

- Forgetting to change the limits of integration

- Making mistakes in the algebraic manipulation of the integral

- Forgetting to replace the new variable with the original variable in the final answer

5. When should I use integration by substitution?

Integration by substitution is most useful when the integral contains a function with a known derivative, such as trigonometric or exponential functions. It is also helpful when the integrand is a complex expression that can be simplified by substitution.

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