Integral using substitution x = -u

[songoku]

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

 

[Office_Shredder]

Is there a specific reason to think that integral is solvable, or that that specific substitution would work?

 

[songoku]

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

 

[Mark44]

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.

 

[songoku]

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

 

[Mark44]

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?

 

[fresh_42]

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.

 

[songoku]

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

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

 

[fresh_42]

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 ...

 

[songoku]

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

 

[songoku]

$$\int_{-a}^a \frac{6x^2+5}{1+2^x}dx=\int_{-a}^0 \frac{6x^2+5}{1+2^x}dx+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=-\int_{a}^0 \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$
$$=\int_{0}^a \frac{6u^2+5}{1+2^{-u}}du+\int_{0}^a \frac{6x^2+5}{1+2^x}dx$$

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

 

[fresh_42]

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.

 

[songoku]

Thank you very much