Definite Integral Problem

[Podrick]

Hi guys

I've been struggling to find the integral of the following function:

x(x^2+7)^12

Is there some sort of chain-rule like formula I could use for questions like this?

Thanks

 

[Deleted member 4993]

Hi guys

I've been struggling to find the integral of the following function:

x(x^2+7)^12

Is there some sort of chain-rule like formula I could use for questions like this?

Thanks

Substitute:

u = x^2 + 7

du = 2x dx

Then

\(\displaystyle \displaystyle{\int x(x^2 + 7)^{12} dx}\)

\(\displaystyle = \ \displaystyle{\frac{1}{2}\int u^{12} du}\)

 

[Podrick]

Thanks for the response!

How did you put the 1/2 out the front though? For a similar question, such as int. (t/(t^2-1)^4)dt, I understand that, by your method, I would let u=t^2-1. Then, du=2t dt.

But I don't understand the next step... Why are we finding the derivative for an integral?

 

[Steven G]

Thanks for the response!

How did you put the 1/2 out the front though? For a similar question, such as int. (t/(t^2-1)^4)dt, I understand that, by your method, I would let u=t^2-1. Then, du=2t dt.

But I don't understand the next step... Why are we finding the derivative for an integral?

you are actually finding du, not du/dx which is the derivative.
You want to write the integral in the form (u^n)du which equals (u^(n+1))/(n+1) +c provided that n is not -1.
You can multiply inside the integral by any number (no x's!) you want/need as long as in front of the integral you multiply by the reciprocal of that number. You get this number from computing du.
In the example you have above, the part of the integral that is (t^2-1)^4 you will replace this with u^4. What is left over? Answer is tdt. Well if you had 2tdt you can replace it with du! So multiply inside the integral by 2 and in front of the integral by 1/2 (note that 2*(1/2)=1). Then you have (1/2) times the integral du/u^4 = (1/2) times the integral (u^-4)du. This integral is easily solved. At the end you replace u with (t^2)-1