How to evaluate the integral $\int_0^1 t\sqrt{4+9t^2} \,dt$ Can somebody please show me how to evaluate the following integral:
$$\int_0^1 t\sqrt{4+9t^2} \,dt?$$
I know how to integrate a square root function, but the $t$ in front throws me off, and if there are associated properties with this sort of integral.
Could you also, knowing that $t=\sqrt{t^2}$, multiply it into the other square root and integrate that way?
 A: Hint: Substitute $u = 4 + 9t^2$, $du = 18 t dt$.
A: Let $\;\color{red}{\bf u = 4 + 9t^2}.\;$ Then $ \,du = 18 t \,dt \iff \color{blue}{\bf t\,dt = \dfrac 1{18}} \,du$.
Now, for our new bounds of integration: when $\;t = 0, \;u = 4 + 9(0)^2 = 9.\;$  When $\;t = 1, \;u = 4+9(1)^2 = 13$.
Now substitute, to get $$\int_0^1 \sqrt{\color{red}{\bf 4 + 9t^2}}\color{blue}{\bf (t\,dt)} \quad = \quad  \color{blue}{\bf \dfrac1{18}}\int_{\bf 4}^{\bf 13} \sqrt{\color{red}{\bf u}} \color{blue}{\bf \,du}\quad  = \quad \dfrac 1{18} \int_4^{13} u^{1/2} \,du$$
Integrate, and then evaluate at the new bounds of integration, so there's no need to back-substitute.
A: Hint:
$$\int f'(x)f^n(x)dx=\frac{f^{n+1}(x)}{n+1}$$
but $f(x)=4+9t^2$ so $f'(x)=18t$
so $$\int t\sqrt{4+9t^2}dx=\frac{1}{18}\int 18t\sqrt{4+9t^2}dx=\frac{1}{27}(\sqrt{(4+9t^2)^3})+C$$
where C is arbitrary constant
A: You say: 

I know how to integrate a square root function

But more exactly, you know how to integrate $$\int u^\frac{1}{2} du$$ So, if you make the thing under the square root sign u,  does the rest of the integrand become $du$?  This is the sort of question that leads, in this case, to the substitution of the responders above...
