Can I not use the substitution method for integrating something like (x^2+1)^(2)?

[The Student]

When I try, it doesn't seem to work. For S(x^2+1)^(2)dx, let u = x^2+1, and du = 2xdx (please excuse the bad looking notation). So Su^(2)du/(2x) = u^3/(6x) = (x^2+1)^3/(6x). But this isn't the right answer. Or am I doing something wrong?

 

[stapel]

When I try, it doesn't seem to work. For S(x^2+1)^(2)dx, let u = x^2+1, and du = 2xdx (please excuse the bad looking notation). So Su^(2)du/(2x) =...

Your integral must be in terms only of "x" or only of "u"; not both.

Instead of substitution, why not just multiply stuff out and integrate, term by term?

. . . . .\(\displaystyle (x^2\, +\, 1)^2\, =\, x^4\, +\, 2x^2\, +\, 1\)

. . . . .\(\displaystyle \displaystyle{ \int\, }\) \(\displaystyle \left(x^4\, +\, 2x^2\, +\, 1\right)\, dx\, =\, \dfrac{x^5}{5}\, +\,...\)

...and so forth.

 

[Steven G]

When I try, it doesn't seem to work. For S(x^2+1)^(2)dx, let u = x^2+1, and du = 2xdx (please excuse the bad looking notation). So Su^(2)du/(2x) = u^3/(6x) = (x^2+1)^3/(6x). But this isn't the right answer. Or am I doing something wrong?

After you make the u-substitution you can not solve the integral until there are no x's. You know that u=x^2+1. Can you solve this for x so that you can get rid of the x in Su^(2)du/(2x)? Will that help??
You seem to treat the x as a constant. So why not do that in the original equation? The answer is that you can't. Just like you can't in the integral Su^(2)du/(2x).

I do not see this integral as an integral that should be solved by u-substitution. If you square x^2 + 1 then you'll have a nice polynomial in x which you should be able to integrate.

Show us how you make out. Good Luck.

 

[The Student]

Thanks for the help Jomo and staple, I get it now.