Integral using substitution x = -u

In summary: Office_Shredder, Mark44 and fresh_42 and I am really sorry for not posting the lower and upper limit of the integralIn summary, the conversation revolved around solving an integral using substitution. The original substitution of ##x=-u## was deemed not useful, and alternative methods were suggested, such as using symmetry and expanding the first quotient by ##2^x##. The conversation ended with a discussion about whether it was valid to change the variable from ##u## back to ##x##, which was deemed acceptable as long as the boundaries were taken into consideration.
  • #1
songoku
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Homework Statement
Using substitution ##x=-u##, solve:

$$\int \frac{6x^2+5}{1+2^x}dx$$
Relevant Equations
Integration by substitution
Integration by parts
Integration by partial fraction
Integration by trigonometry substitution
Is it possible to solve this integral? I think the substitution ##x=-u## does not help at all since it only changes variable ##x## to ##u## without changing the integrand much.

Using that substitution:
$$\int \frac{6x^2+5}{1+2^x}dx=-\int \frac{6u^2+5}{1+2^{-u}}du$$

Then how to continue?

Thanks
 
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  • #2
Is there a specific reason to think that integral is solvable, or that that specific substitution would work?
 
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  • #3
Office_Shredder said:
Is there a specific reason to think that integral is solvable, or that that specific substitution would work?
I don't think so, at least that's my opinion. I posted the full and exact question I got so substitution ##x=-u## is actually given by the question
 
  • #4
songoku said:
I don't think so, at least that's my opinion. I posted the full and exact question I got so substitution ##x=-u## is actually given by the question
Really? This is the exact wording of the question?
Using substitution ##x=-u##, solve:

$$\int \frac{6x^2+5}{1+2^x}dx$$
I agree that this is not a useful substitution for this problem, and I can't think of any substitution that is likely to be successful.
 
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  • #5
Mark44 said:
Really? This is the exact wording of the question?
Actually the question is about definite integral so there is lower limit ##-a## and upper limit ##a## where ##a>0## but I think the limit is not really important so I just omit it at first.

Is the limit actually important to solve the integral? If yes, I am really sorry for omitting it
 
  • #6
songoku said:
Actually the question is about definite integral so there is lower limit ##-a## and upper limit ##a## where ##a>0## but I think the limit is not really important so I just omit it at first.

Is the limit actually important to solve the integral? If yes, I am really sorry for omitting it
I think that the limits of integration are significant, but I don't see how they help us here. Are there any similar examples in your textbook?
 
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  • #7
Think about symmetries. ##\int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_{0}^a f(x)\,dx##. Now perform the substitution on the first integral and leave the second as is.
 
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  • #8
Mark44 said:
I think that the limits of integration are significant, but I don't see how they help us here. Are there any similar examples in your textbook?
I have no textbook and there is no similar example given previously

fresh_42 said:
Think about symmetries. ##\int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_{0}^a f(x)\,dx##. Now perform the substitution on the first integral and leave the second as is.
$$\int_{-a}^a \frac{6x^2+5}{1+2^x}dx=\int_{-a}^0 \frac{6x^2+5}{1+2^x}dx+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=-\int_{a}^0 \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=\int_{0}^a \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$

Then how to continue?

Thanks
 
  • #9
I haven't solved it for you. That was just an idea. Now we have
$$
\int_0^a \dfrac{6x^2+5}{1+2^{-x}}+\dfrac{6x^2+5}{1+2^{x}}\,dx
$$
Next, expand the first quotient by ##2^x## and add the fractions, then ...
 
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  • #10
fresh_42 said:
I haven't solved it for you. That was just an idea. Now we have
$$
\int_0^a \dfrac{6x^2+5}{1+2^{-x}}+\dfrac{6x^2+5}{1+2^{x}}\,dx
$$
Next, expand the first quotient by ##2^x## and add the fractions, then ...
I think I understand your hint

Thank you very much Office_Shredder, Mark44 and fresh_42 and I am really sorry for not posting the lower and upper limit of the integral
 
  • #11
songoku said:
$$\int_{-a}^a \frac{6x^2+5}{1+2^x}dx=\int_{-a}^0 \frac{6x^2+5}{1+2^x}dx+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=-\int_{a}^0 \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=\int_{0}^a \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
fresh_42 said:
I haven't solved it for you. That was just an idea. Now we have
$$
\int_0^a \dfrac{6x^2+5}{1+2^{-x}}+\dfrac{6x^2+5}{1+2^{x}}\,dx
$$
Next, expand the first quotient by ##2^x## and add the fractions, then ...
Sorry @fresh_42, I want to ask again about this part.

In my last line, I got:
$$=\int_{0}^a \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$

But then you changed it into:
$$
\int_0^a \dfrac{6x^2+5}{1+2^{-x}}+\dfrac{6x^2+5}{1+2^{x}}\,dx
$$

Can we do that, just change ##u## into ##x##? Because in the beginning we use substitution ##x=-u## so changing ##u## to ##x## seems contradicts the initial substitution

Thanks
 
  • #12
songoku said:
Can we do that, just change ##u## into ##x##? Because in the beginning we use substitution ##x=-u## so changing ##u## to ##x## seems contradicts the initial substitution

Thanks
##\int f(x)dx = \int f(u)du## so all we have to be careful with are the boundaries! Whether we write an integral with ##u,## ##x,## or any other letter doesn't matter.
 
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  • #13
Thank you very much
 

1. What is the purpose of using substitution x = -u in an integral?

Substitution x = -u is used to simplify integrals that involve functions with negative exponents or radicals. It allows us to transform the integral into a more manageable form.

2. How do you perform substitution x = -u in an integral?

To perform substitution x = -u in an integral, you need to replace all instances of x with -u and also change the limits of integration accordingly. This will result in a new integral that can be solved using standard techniques.

3. Can you explain the concept of "u-substitution" in integrals?

U-substitution, also known as the substitution method, is a technique used to simplify integrals by substituting a variable with a new variable. This new variable, usually represented by u, helps to transform the integral into a more manageable form.

4. What are some common mistakes to avoid when using substitution x = -u in an integral?

Some common mistakes to avoid when using substitution x = -u in an integral include forgetting to change the limits of integration, not substituting all instances of x, and not simplifying the integral after substitution.

5. Can substitution x = -u be used in all integrals?

No, substitution x = -u can only be used in integrals where the variable x appears with a negative exponent or under a radical. It is not applicable to all integrals and other methods, such as integration by parts, may be needed to solve the integral.

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