evalute $\displaystyle \int_{0}^{1} \dfrac{\sqrt{t}}{t+1}dt$
$\displaystyle u = \sqrt{t}, u^2 = t, 2udu = dt $
need help setting this up:
$\displaystyle \int \frac{u*2udu}{u^2+1}$
$\displaystyle \frac{\sqrt{x}}{x+1} = \frac{x}{\sqrt{x} (x+1)} = \frac{2x}{\boxed{2\sqrt{x}} (x+1)}$
The thing in the box is the derivative, that is why I wrote it like that if I use the substitution:
$\displaystyle t = \sqrt{x} \Rightarrow t' = \frac{1}{2\sqrt{x}}$
That means,
$\displaystyle \int_0^1 \frac{t^2}{t^2+1} dt$
Can you do if frum heir?