# Thread: integral with square root

1. ## integral with square root

Hi everyone,

In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

Int. 0-1 r^3/sq. 4+r^2 dr

t=r^3
tu=3r^2dr
r^2dr=tu/3

Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

Thank you

2. Set $\displaystyle u=\sqrt{4+r^2}$

3. You better use the substitution
$\displaystyle \sqrt{4+r^2}=u\Rightarrow 4+r^2=u^2\Rightarrow rdr=udu$
$\displaystyle r=0\Rightarrow u=2$
$\displaystyle r=1\Rightarrow u=\sqrt{5}$
Now
$\displaystyle \displaystyle\int_0^1r^2\sqrt{4+r^2}\cdot rdr=\int_2^{\sqrt{5}}(u^2-4)u^2du$
and you can continue.

4. Hi,

Can you explain how you got "" please?

This is what I did:

u squared -4=(2 squared-4)-(sq. 5) sq. -4=-1

Thank you

5. $\displaystyle u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr$, so $\displaystyle u\,du=r\,dr$

Does that make sense?

6. Yes, thank you very much

7. Originally Posted by Krizalid
$\displaystyle u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr$, so $\displaystyle u\,du=r\,dr$

Does that make sense?

8. Originally Posted by chocolatelover
Hi everyone,

In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

Int. 0-1 r^3/sq. 4+r^2 dr

t=r^3
tu=3r^2dr
r^2dr=tu/3

Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

Thank you
sssss