Casey's First Day: Solving $\int t \sqrt{7t^2+12}dt$

[Saladsamurai]

First day of u subs...

$$\int t \sqrt{7t^2+12}dt$$

I am assuming that u=t, but It is maiking a mess when I do that.

Just a hint please,
Casey

 

[bob1182006]

your first choice on a u-substitution with a rational number should always be the entire thing under the square root.

try u=7*t^2+12 instead

 

[rocomath]

let u be the radican (is that the proper term? i forget) :D

 

[Saladsamurai]

let u be the radican (is that the proper term? i forget) :D

do you mean like this?

your first choice on a u-substitution with a rational number should always be the entire thing under the square root.

try u=7*t^2+12 instead

If I do this, I get $$\int tudt$$ and $$du=\frac{dt}{2\sqrt{7t^2+12}}$$ ...right?

I think I am confused...

 

[bob1182006]

no just the 7t^2+12

if u=7t^2+12
what is dt=??

 

[Saladsamurai]

Oh..one sec...

 

[Saladsamurai]

Brain Cramp!$$u=7t^2+12$$
so
$$du=14tdt$$
$$\int t u^{1/2} dt *14*\frac{1}{14}$$
$$=\frac{1}{14}\int \sqrt{u}* du$$
and I got it from here..
Thanks guys,
Casey