# Thread: Integration.

1. ## Integration.

I believe what's being shown is a method of finding an integral called "u substitution". So I understand that $\displaystyle u=x^4+2$. Also I understand that $\displaystyle du=4x^3dx$ although I wonder at this point why is a derivative being taken? I thought we were finding the anti derivative? Following further I understand that [tex](x^4+2)^5=u^5. Then there is $\displaystyle x^3dx=\frac{1}{4}du$ which I don't understand at all. I am just left wondering where the $\displaystyle \frac{1}{4}$ came from.

At this point in my online course they are talking about the anti derivative and we have yet only talked about the indefinite integral.

Thanks for any responses...

2. ## Re: Integration.

you need to find the formula for du in terms of dx so that you can replace dx in the original integral by a function of du.

Here we have

u = x4 + 2

du/dx = 4x3

du = 4x3 dx

1/4 du = x3 dx

the right hand side is what appears in your original integral.

So after substituting you end up with

$\displaystyle \int {u^5} \frac{1}{4}\,du = \frac{1}{4}\ \int u^5 \,du$

3. ## Re: Integration.

I know that this opinion is in the vast minority but, I always discouraged students from using u-substitution.
I think it is pathetically sad if a student sees $\displaystyle \int {{x^3}{{\left( {{x^4} + 2} \right)}^5}dx}$ and does not see at once that the answer is $\displaystyle \frac{1}{{24}}{\left( {{x^4} + 2} \right)^6}$.

For years I used such tests questions as: Prove that $\displaystyle \int {{x^3}{{\left( {{x^4} + 2} \right)}^5}dx}=\frac{1}{{24}}{\left( {{x^4} + 2} \right)^6}+C$.
Students were often floored to see that only differentiation is needed as a proof.
But it always got my point across: Learn the differentiation forms.