# Math Help - Simple antideriv.

1. ## Simple antideriv.

I am wondering if I'm overlooking a simple way of solving this, without expanding the squared term.

$\int 3x(x+1)^2\,dx$

If I have to first expand, that's fine. I was just thinking I can do it without.

2. Originally Posted by scorpion007
I am wondering if I'm overlooking a simple way of solving this, without expanding the squared term.

$\int 3x(x+1)^2\,dx$

If I have to first expand, that's fine. I was just thinking I can do it without.
You can do it without expanding, but I'm not sure it will be simpler. You could do this by parts, or change the variable.

CB

3. Originally Posted by scorpion007
I am wondering if I'm overlooking a simple way of solving this, without expanding the squared term.

$\int 3x(x+1)^2\,dx$

If I have to first expand, that's fine. I was just thinking I can do it without.
You could u-substitute, or do integration by parts. But I think that just expanding would be by far the easiest approach.

4. Would u-substitution really work here? If I let u=x+1, then du=dx ... and there's no part of du in the original integrand.

5. Originally Posted by scorpion007
Would u-substitution really work here? If I let u=x+1, then du=dx ... and there's no part of du in the original integrand.
You could re-substitute for the $3x$, using $u=x+1$, i.e. $x=u-1$. Though it definitely would not be the best route in my opinion.

6. ah, I see. I always forget about that step.

Thanks anyhow for the advice.

7. Originally Posted by scorpion007
I am wondering if I'm overlooking a simple way of solving this, without expanding the squared term.

$\int 3x(x+1)^2\,dx$

If I have to first expand, that's fine. I was just thinking I can do it without.
Do it the easiest way that you can (which in this case is probably by expanding), when you are no longer studying this stuff you may as well just use technology to do these things; there is little intrinsic merit in remembering a bag of tricks for systematic integration (because you will never be as good at it as a machine implementing the Risch algorithm).

The only area in which you may be better off on your own is knowing which integrals can be expressed in terms of higher transcendental function after some rearrangement.

CB