1. ## rationalizing substitution

evalute $\int_{0}^{1} \dfrac{\sqrt{t}}{t+1}dt$

$u = \sqrt{t}, u^2 = t, 2udu = dt$

need help setting this up:
$\int \frac{u*2udu}{u^2+1}$

2. i think i got the answer: $2\sqrt{1} -2arctan(\sqrt{1}) - 2\sqrt{0}-2arctan(\sqrt{0})$

3. Originally Posted by viet
evalute $\int_{0}^{1} \dfrac{\sqrt{t}}{t+1}dt$

$u = \sqrt{t}, u^2 = t, 2udu = dt$

need help setting this up:
$\int \frac{u*2udu}{u^2+1}$
$\frac{\sqrt{x}}{x+1} = \frac{x}{\sqrt{x} (x+1)} = \frac{2x}{\boxed{2\sqrt{x}} (x+1)}$

The thing in the box is the derivative, that is why I wrote it like that if I use the substitution:
$t = \sqrt{x} \Rightarrow t' = \frac{1}{2\sqrt{x}}$

That means,
$\int_0^1 \frac{t^2}{t^2+1} dt$

Can you do if frum heir?

4. yea thanks

5. ## This is the substitution

Originally Posted by viet
evalute $\int_{0}^{1} \dfrac{\sqrt{t}}{t+1}dt$

$u = \sqrt{t}, u^2 = t, 2udu = dt$

need help setting this up:
$\int \frac{u*2udu}{u^2+1}$
This is the substitution: