# Integration help

James889
Hi,

I would like some help with the following integral

$$\int t^3\cdot \sqrt{1+t^2}~dt$$

I don't know what to do with the $$t^3$$ term.
If it wasn't there it would be easy.

Help appreciated.

## Answers and Replies

benhou
how about letting $$u=1+t^{2}$$?

James889
how about letting $$u=1+t^{2}$$?

Thats what i thought about too, at first.

But it doesn't get me anywhere

$$t^3\cdot\sqrt{u}$$

benhou
Well, before taking the integral, you have to have only one variable. What you need to do is to represent dt in terms of du, and t^3 in terms of u.

James889
Well, before taking the integral, you have to have only one variable. What you need to do is to represent dt in terms of du, and t^3 in terms of u.

Aha!
I think i understand.

So i need $$\int(u-1)\sqrt{u}du$$

How do i proceed now?

benhou
Well, I think you are done. Don't you see it yet? Just expand and integrate.

benhou
You made a mistake. since
$$u=1+t^{2}$$
$$du=2tdt$$ or
$$dt=du/(2t)$$

You should have 1/2 somewhere.

Last edited:
James889
You made a mistake. since
$$u=1+t^{2}$$
$$du=2tdt$$ or
$$dt=du/(2t)$$

You should have 1/2 somewhere.

Oh, dang i forgot about that. Thanks!

$$\frac{1}{5}u^{5/2} -u^{3/2}$$

benhou
Don't forget 1/3